A) The electric field inside the paint layer is zero
B) The electric field just outside the paint layer is [tex]3.2\cdot 10^7 N/C[/tex] (radially inward)
C) The electric field at 6.00 cm from the surface is [tex]1.2\cdot 10^7 N/C[/tex] (radially inward)
Explanation:
A)
We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:
[tex]\int EdS = \frac{q}{\epsilon_0}[/tex]
where
E is the magnitude of the electric field
dS is the element of the surface
q is the charge contained within the surface
[tex]\epsilon_0[/tex] is the vacuum permittivity
By taking a sphere centered in the origin,
[tex]\int E dS = E \cdot 4\pi r^2[/tex]
where [tex]4\pi r^2[/tex] is the surface of the Gaussian sphere of radius r.
In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than
[tex]R=9.0 cm = 0.09 m[/tex] (radius of the plastic sphere is half of the diameter)
Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:
[tex]q=0[/tex]
And therefore,
[tex]E4\pi r^2 = 0\\\rightarrow E = 0[/tex]
So, the electric field inside the plastic sphere is zero.
B)
Here we apply again Gauss Law:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so
[tex]r=R=0.18 m[/tex]
The charge contained within the Gaussian sphere is therefore
[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]
Therefore, the electric field is
[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C[/tex]
And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is
[tex]E=3.2\cdot 10^7 N/C[/tex]
C)
For this part again, we apply Gauss Law:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be
r = 9 cm + 6 cm = 15 cm = 0.15 m
While the charge contained within the sphere is again
[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]
Therefore, the electric field in this case is
[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C[/tex]
And again, this is radially inward, so according to the sign convention asked in the problem,
[tex]E=1.2\cdot 10^7 N/C[/tex]
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To find the electric field just inside the paint layer, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.
A) Inside the paint layer:
Since the paint is spread in a very thin uniform layer over the surface of the plastic sphere, we can consider a Gaussian surface just inside the paint layer, with a radius slightly smaller than the sphere's radius (let's call it r). The charge enclosed within this Gaussian surface is the charge on the sphere, which is -29.0 μC.
Gauss's law equation can be written as:
Φ = Q_enclosed / ε₀
The electric field just inside the paint layer (E₁) is given by:
E₁ * A = Q_enclosed / ε₀
E₁ * 4πr² = -29.0 μC / ε₀
E₁ = (-29.0 μC / ε₀) / (4πr²)
The direction of the electric field just inside the paint layer will be radially outward, opposite to the direction of the charge.
B) Just outside the paint layer:
The electric field just outside the paint layer (E₂) will be the same as the electric field inside the sphere when it was uncharged. This is because the charge on the paint layer is spread on the outer surface of the sphere.
So, E₂ = E₁ (since it's radially outward)
C) 6.00 cm outside the surface of the paint layer:
To find the electric field 6.00 cm outside the surface of the paint layer, we need to calculate the electric field due to the charge on the sphere.
E₃ = k * Q / r²
where k is the Coulomb's constant (k = 9.0 x 10^9 N m²/C²), Q is the charge on the sphere (-29.0 μC), and r is the distance from the center of the sphere to the point where we want to find the electric field (6.00 cm = 0.06 m + radius of the sphere).
Remember that the electric field due to a charged object decreases with the square of the distance.
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An astronaut is in space with a baseball and a bowling balL The astronaut gives both objects an equal push in the samedirection. Does the baseball have the same inertia as the bowling ball? Why? Does the baseball have the sameacceleration as the bowling ball from the push? Why? If both balls are traveling at the same speed, does the baseball havethe same momentum as the bowling ball?
For all solutions the answer is NO. And this is easily intuited because for the three conditions there is the dependence of the mass against some physical property of movement. Both bodies do not have the same mass.
In the case of Inertia, it is understood that it is the tendency of an object to resist change and is mass dependent. The object with greater mass will tend to resist change. Since the mass of the bowling ball is greater than the base ball, the bowling ball has greater inertia compared to the base ball.
For the second part, remember that force, according to Newton's second law, is defined as the product between mass and acceleration, so the bowling ball will accelerate less by having a greater mass.
Finally, momentum is defined as the product between mass and velocity. The mass is greater than one of its objects even though the speeds are the same. Therefore, the momentum of the bowling ball is greater than the momentum of baseball.
In space, the baseball has less inertia, greater acceleration, and less momentum compared to the bowling ball when both receive an equal push, due to their differences in mass.
The question relates to the concept of inertia, acceleration, and momentum in physics. When the astronaut in space gives an equal push to a baseball and a bowling ball, the baseball does not have the same inertia as the bowling ball, because inertia is a measure of an object's resistance to changes in its state of motion and is directly proportional to the object's mass. Since the bowling ball has a greater mass than the baseball, it has more inertia.
As a result, the baseball will have a greater acceleration than the bowling ball from the same force, due to Newton's second law of motion (F=ma), which states that force equals mass times acceleration. If we consider both balls traveling at the same speed, the baseball will not have the same momentum as the bowling ball, because momentum is the product of mass and velocity (p=mv), and the bowling ball has a greater mass.
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field zero?
Final answer:
To find where the electric field is zero between two point charges (5.0 μC and -4.0 μC), set up an equation based on Coulomb's Law and solve it considering that the point lies closer to the smaller magnitude charge.
Explanation:
The question requires the application of concepts from electrostatics, specifically the properties of the electric field generated by point charges. To determine the point at which the electric field is zero, one must consider the magnitudes of the charges and their distances from the point of interest. Since electric fields generated by individual charges superpose, the point where the electric field is zero is where the electric field due to one charge balances out the electric field due to the other charge.
For two charges Q1 (+5.0 μC) and Q2 (-4.0 μC) separated by 50 cm (0.5 meters), the electric field is zero at a point that is closer to the smaller magnitude charge. Let's call the distance from Q1 to the point where the field is zero 'd'. Using Coulomb's Law and the concept of superposition, we can set up the equation:
E1 = E2
|k * Q1 / d^2| = |k * Q2 / (0.5 - d)^2|
|Q1/d^2| = |Q2/(0.5 - d)^2|
Substituting Q1 and Q2 with their respective values, we can solve for 'd' using algebraic methods to find the point on the meter stick where the electric field equals zero. However, since the solution requires calculations, it is important to apply the proper mathematical steps to reach the correct conclusion.
A 0.1 m by 0.1 m sheet of cardboard is placed in a uniform electric field of 10 N/C. At first, the plane of the sheet is oriented perpendicular to the electric field vector so that the electric flux through the sheet is 0.01 N-m2/C. By what angle do you need to rotate the sheet to reduce the electric flux by 1/2?
Answer:
The angle is 89°.
Explanation:
Given that,
Electric field = 10 N/C
Electric flux = 0.01 N-m²/C
Area [tex]A=\pi\times(0.1)^2[/tex]
We need to calculate the angle
Using formula of electric flux
[tex]\phi=EA\cos\theta[/tex]
[tex]\cos\theta=\dfrac{\phi}{EA}[/tex]
Where, E = electric field
[tex]\phi[/tex] = electric flux
A = area
Put the value into the formula
[tex] \cos\theta=\dfrac{\dfrac{0.01}{2}}{10\times\pi\times(0.1)^2}[/tex]
[tex]\theta=\cos^{-1}(0.01592)[/tex]
[tex]\theta=89.0^{\circ}[/tex]
Hence, The angle is 89°.
The angle required to rotate the sheet to reduce the electric flux by 1/2 is 89 degrees.
What is electric flux?The number of electric lines that interact the area of a given object or space.
It can be given as,
[tex]\phi=ES\cos \theta[/tex]
Here, [tex]E[/tex] is the magnitude of electric field, [tex]S[/tex] is the area of surface and [tex]\theta[/tex] is the angle between electric field and perpendicular.
Given information-
The dimensions of sheet of cardboard is 0.1 by 0.1.
The sheet is placed between the uniform electricity field of 10 N/C.
Put the values in the above formula as,
[tex]\dfrac{0.01}{2} =10\times\pi\times{0.1^2}\times\cos \theta\\\theta=cos^-(0.01592)\\\theta=89^o[/tex]
Hence the angle required to rotate the sheet to reduce the electric flux by 1/2 is 89 degrees.
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A 20 μF capacitor has previously been charged up to contain a total charge of Q=100 μC on it. The capacitor is then discharged by connecting it directly across a 100kΩ resistor. Calculate the charge remaining on the capacitor exactly 3.00 seconds after being connected to the resistor.
Answer:
Q= 22.3 μC
Explanation:
Given that
C= 20 μF
Qo= 100 μC
R= 100 kΩ
t= 3 s
T= R C
T= 100 x 1000 x 20 x 10⁻⁶ s
T=2 s
We know that charge on the capacitor is given as
[tex]Q=Q_0e^{\dfrac{-t}{T}}[/tex]
[tex]Q=100\times 10^{-6}\times e^{\dfrac{-3}{2}}[/tex]
Q= 0.0000223 C
Q= 22.3 μC
Final answer:
The charge remaining on the capacitor after 3 seconds of discharging is approximately 22.31 µC, calculated using the formula for exponential decay of charge in an RC circuit.
Explanation:
To calculate the charge remaining on a 20 µF capacitor after 3 seconds of discharging through a 100 kΩ resistor, we'll use the formula for exponential decay of charge in an RC circuit, which is Q(t) = Q0 e-(t/RC), where Q0 is the initial charge, t is the time, and RC is the time constant of the circuit. The time constant (RC) for this circuit can be calculated as (100 × 103 Ω)(20 × 10-6 F) = 2 seconds. Plugging in the given values, we find that Q(3s) = 100 µC × e-(3/2), which after calculation gives a remaining charge of approximately 22.31 µC.
A 4.89 μC test charge is placed 4.10 cm away from a large, flat, uniformly charged nonconducting surface. The force on the charge is 321 N. The charge is then moved 2.00 cm farther away from the surface. What is the force on the test charge now?
Final answer:
The force on the test charge remains the same at 321 N after it is moved farther away because the electric field produced by a large, flat, uniformly charged surface is constant close to the surface.
Explanation:
The question involves calculating the new force on a test charge after it is moved farther away from a charged surface. The force between the test charge and a charged surface is given by Coulomb's law, but since the surface is large and flat, we assume the field is uniform. Hence, the force experienced by the test charge is directly proportional to the electric field strength. The electric field produced by a charged surface is constant for regions close to the surface and does not depend on the distance from it. Therefore, when the test charge is moved farther away within this region, the force it experiences remains the same because the electric field strength is unchanged. In this scenario, after moving the charge from 4.10 cm to a new distance of 6.10 cm (an additional 2.00 cm), the force on the test charge will remain at 321 N.
Many satellites orbit Earth at maximum altitudes above Earth's surface of 1000 km or less. Geosynchronous satellites, however, orbit at an altitude of 35790 km above Earth's surface. How much more energy is required to launch a 410 kg satellite into a geosynchronous orbit than into an orbit 1000 km above the surface of Earth?
Answer:
6.26 times more
Explanation:
Given:
- most satellite orbit at height r_1 = 1000 km
- Geosynchronous satellites orbit at height r_2 = 35,790 km
- mass of Geosynchronous satellite m = 410 kg
- The radius of the earth r_e = 6371 km
Find:
- Compare the Energy required to send the satellite to Geosynchronous orbit @ r_2 vs Energy required to send the satellite to normal orbit @ r_1. How much more. ( U_1 / U_2 ).
Solution:
- The gravitational potential energy of any mass m in an orbit around another mass M is given by the following relation:
U_g = - G*m*M / r
Where,
G : Gravitational constant
- We compute the gravitational potential energy U_g of the satellite at both orbits as follows:
-Normal orbit U_1 = - G*m*M / r_e + G*m*M / (r_e+r_1)
U_2 = - G*m*M / r_e + G*m*M / (r_2+r_e)
Now: Take a ratio of the two energies U_1 and U_2 as follows:
U_2 / U_1 = (- G*m*M / r_e + G*m*M / r_2+r_e) / (- G*m*M / r_e + G*m*M / r_1+r_e)
U_2 / U_1 = (1 / (r_2+r_e) - 1 / r_e ) / (1 / (r_1 + r_e) - 1 / r_e )
- Plug values:
U_2 / U_1 = (1 / (35790+6371) - 1 / 6371 ) / (1 / (1000+6371) - 1 / 6371 )
- Evaluate:
U_2 / U_1 = (-1.33242681 * 10^-4) / (2.12944*10^-5)
U_2 / U_1 = 6.26
- Hence The energy required to send the satellite to Geosynchronous orbit is 6.26 times more than that required for normal orbit.
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravity") on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and an astronaut is strapped in at the other end. Suppose that he is aligned along the centrifuge’s arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this device is typically 12.5 g. (a) How fast must the astronaut’s head be moving to experience this maximum acceleration? (b) What is the difference between the acceleration of his head and feet if the astronaut is 2.00 m tall? (c) How fast in rpm (rev/min) is the arm turning to produce the maximum sustained acceleration?
Answer:
32.9242311983 m/s
1.40506567727g
35.3405770759 rpm
Explanation:
v = Linear Velocity of the capsule
[tex]a_c[/tex] = Centripetal acceleration = [tex]12.5g=12.5\times 9.81[/tex]
r = Radius of the centrifuge = 8.84 m
l = Person's height = 2 m
Centripetal acceleration is given by
[tex]a_c=\frac{v^2}{r}\\\Rightarrow v=a_cr\\\Rightarrow v=\sqrt{12.5\times 9.81\times 8.84}[/tex]
The linear speed of the capsule is 32.9242311983 m/s
The radius would be
[tex]r=\sqrt{r^2+\dfrac{l^2}{4}}\\\Rightarrow r=\sqrt{8.84^2+\dfrac{2^2}{4}}\\\Rightarrow r=8.89638128679\ m[/tex]
The centripetal acceleration
[tex]a_{c2}=\dfrac{32.9242311983^2}{8.89638128679\times 9.81}g\\\Rightarrow a_{c2}=12.4207805891g[/tex]
Change in acceleration from Pythagoras law
[tex]a=\sqrt{a_{ch}^2-a_{c2}^2}\\\Rightarrow a=\sqrt{12.5^2g^2-12.4207805891^2g^2}\\\Rightarrow a=1.40506567727g[/tex]
The difference is 1.40506567727g
Velocity
[tex]v=\omega r\\\Rightarrow v=2\pi Nr\\\Rightarrow N=\dfrac{v}{2\pi r}\\\Rightarrow N=\dfrac{32.9242311983}{2\pi 8.89638128679}\\\Rightarrow N=0.589009617932\ rev/s[/tex]
[tex]N=0.589009617932\times 60=35.3405770759\ rpm[/tex]
The speed is 35.3405770759 rpm
A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.9 km , the jet is moving with a speed of 400 m/s.
a. What is the jet's acceleration, assuming it to be a constant acceleration?
b-Is your answer reasonable ? Explain.
The jet's acceleration is calculated as 7.14 m/s² using a kinematic equation. This value is reasonable for a jet plane given its powerful engines. High-speed aircraft often have high accelerations.
To determine the jet's acceleration, we can apply the kinematic equation:
vf² = vi² + 2aΔx
where:
vf is the final velocity (400 m/s)vi is the initial velocity (300 m/s)a is the accelerationΔx is the displacement (4.9 km or 4900 m)Solving for acceleration a:
400² = 300² + 2a(4900)
160000 = 90000 + 9800a
70000 = 9800a
a = 7.14 m/s²
Reasonableness of the Answer
Yes, the answer is reasonable. Accelerations for jet planes are typically high due to the powerful engines they possess. An acceleration of 7.14 m/s² aligns well with the capabilities of high-speed aircraft.
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 4.0 cm . Two of the particles have a negative charge: q1
Answer:
The question continues ; Two of the particles have a negative charge: q1 = -6.3nC and q2 = -12.6nC . The remaining particle has a positive charge, q3 = 8.0nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
Explanation:
The step by step and mathematical interpretation is as shown in the attached file.
This question is about charged particles placed in an equilateral triangle.
The subject of this question is Physics and it is at a High School level.
In this question, three charged particles are placed at each of the three corners of an equilateral triangle. The sides of the triangle are given to be 4.0 cm long. Two of the particles have a negative charge, which is denoted as q1.
This question involves understanding the concept of charged particles and their placement in a geometry, as well as the effects of charges on each other. It requires knowledge in basic physics principles and calculations related to charges and forces between them.
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Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he swims the first 100 meters at an average speed of 1.80 m/s how fast must he swim the second 100 meters in order to break the record?
Answer:
Explanation:
Given
average speed of Phelps [tex]v_{avg}=2\ m/s[/tex]
total distance [tex]d=200\ m[/tex]
he swims first 100 m at an average speed if [tex]1.8 m/s[/tex]
so time taken is [tex]t_1=\frac{100}{1.8}=55.55\ s[/tex]
suppose [tex]t_2[/tex] is the time taken to swim remaining half
average velocity is [tex]v_{avg}=\frac{displacement}{total\ time}[/tex]
[tex]v_{avg}=\frac{100+100}{55.55+t_2}[/tex]
[tex]t_2+55.55=\frac{200}{2}=100[/tex]
[tex]t_2=44.45\ s[/tex]
so velocity in the second half is
[tex]v_2=\frac{100}{45.45}[/tex]
[tex]v_2=2.19\approx 2.2\ m/s[/tex]
Final answer:
Michael Phelps must swim the second 100 meters at an average speed of 2.25 m/s to break the world record in the 200 m freestyle.
Explanation:
To calculate how fast Michael Phelps must swim the second 100 meters to break the world record, let's use the formula for average speed, which is total distance divided by total time. First, we find the time it takes for him to swim the first 100 meters at 1.80 m/s, which is 100 m / 1.80 m/s = 55.56 seconds (rounded to two decimal places). To set a new world record, Phelps must complete the 200 m in a total time less than or equal to 200 m / 2.00 m/s = 100 seconds. Thus, for the second 100 meters, he has 100 seconds - 55.56 seconds = 44.44 seconds. The speed required for the second 100 meters is then 100 m / 44.44 s, which equals 2.25 m/s.
A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pressure and temperature 300 K. Find (a) the mass of the gas, (b) the gravitational force exerted on it, and (c) the force it exerts on each face of the cube. (d) Why does such a small sample exert such a great force? (6%)
Answer:
a) m = 1.174 grams
b) F_g = 0.01151 N
c) F_c = 1013 N
Explanation:
Given:
- The length of a cube L = 10.0 cm
- The molar mass of air M = 28.9 g/mol
- Pressure of air P = 101.3 KPa
- Temperature of air T = 300 K
- Universal Gas constant R = 8.314 J/kgK
Find:
(a) the mass of the gas
(b) the gravitational force exerted on it
(c) the force it exerts on each face of the cube
(d) Why does such a small sample exert such a great force? (6%)
Solution:
- Compute the volume of the cube:
V = L^3 = 0.1^3 = 0.001 m^3
- Use Ideal gas law equation and compute number of moles of air n:
P*V = n*R*T
n = P*V / R*T
n = 101.3*10^3 * 0.001 / 8.314*300
n = 0.04061 moles
- Compute the mass of the gas:
m = n*M
m = 0.04061*28.9
m = 1.174 grams
- The gravitational force exerted on the mass of gas is due to its weight:
F_g = m*g
F_g = 1.174*9.81*10^-3
F_g = 0.01151 N
- The force exerted on each face of cube is due its surface area:
F_c = P*A
F_c = (101.3*10^3)*(0.1)^2
F_c = 1013 N
- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.
(a) The mass of the gas is 1.174g
(b) Gravitational force on it is 0.0115N
(c) Force exerted on the container walls is 1.01×10³N
(d) High kinetic energy of the molecules
Ideal gas in a container:The sides of the cubical container are, L = 10cm = 0.1m
Pressure of the gas, P = 1 atm = 1.01×10⁵ Pa
Temperature of the gas, T = 300K
Molas mass, M = 28.9 g/mol
(a) The volume of the container:
V = L³ = 0.1³
V = 0.001m³
From the idea gas equation:
PV = nRT
n = PV/RT
n = (1.01×10⁵)(0.001)/8.314×300
n = 0.04061 moles
So, the mass of the gas is:
m = nM = 0.04061×28.9
m = 1.174g
(b) the gravitaitonal force on the gas is given by:
f = mg
f = 1.174×10⁻³×9.8
f = 0.0115N
(c) the force exerted on each wall of the cube is :
F = PA
where A is the area of each wall
F = 1.01×10⁵×0.1×0.1
F = 1.01×10³ N
(d) The atoms/molecules of the gas have high kinetic energies, they constantly collide with the walls of the container and transfer large momentum which results in such large force.
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Vector A points in the negative y-direction and has a magnitude of 14 units. Vector B has twice the magnitude and points in the positive x-direction.
(a) Find the direction and magnitude of A + B. (degrees counterclockwise from the +x axis)
(b) |A + B| = ?
(c) Find the direction and magnitude of A - B. (degrees counterclockwise from the +x axis)
(d) |A - B| =?
(e) Find the direction and magnitude of B - A. (degrees counterclockwise from the +x axis)
To find the magnitude and direction of vectors A+B, A-B, and B-A, we perform vector addition and subtraction and calculate the magnitudes using the Pythagorean theorem. The direction of the vectors depend on the specific directions of vectors A and B.
Explanation:The subject of this question is in the field of Physics, specifically under vectors. The grade level of this question falls under High School.
(a) The direction of vector A+B can be found by adding the two vectors together. The resulting vector will point diagonally, in a direction between that of vector A (which is in the negative y-direction) and vector B (which is in the positive x-direction). The magnitude of the resulting vector is √((14)^2 + (2*14)^2)= 24.5
(b) The magnitude of vector A+B is also known as the modulus of the vector, and it is the same as the magnitude calculated in part (a), which is 24.5 units.
(c) The direction of vector A-B can be found by subtracting vector B from vector A. The direction will be in the negative x-direction. The magnitude of the resulting vector is √((14)^2 + (2*14)^2) = 24.5
(d) The magnitude of the vector A-B, also known as the modulus of the vector, is given by the same calculation as in part (c), which is 24.5 units.
(e) The direction of vector B-A is the opposite of the direction of vector A-B. Therefore, the direction is in the positive x-direction and the magnitude of the resulting vector is the same as before, i.e., 24.5 units.
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What are the largest optical telescopes in use today? Why do astronomers want their telescopes to be as large as possible?
The largest telescope currently used is the Gran Telescopio Canarias, (also known as GTC or GRANTECAN). It is 10.4m in diameter, slightly larger than the Keck telescopes in Hawaii.
The telescope observes the visible and infrared light coming from space and has a primary mirror of 10.4 meters, segmented into 36 hexagonal glass-ceramic pieces, 1.9 m between vertices, 8 cm thick, and 470 kg of mass each. The optical system is completed with two mirrors (secondary and tertiary) that form an image in seven focal stations.
Optically the diameter directly influences the magnification of the image. This added to the fact that astronomical objects are quite far away, a telescope of this magnitude allows to obtain more precise images of what is observed in space
Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550 kPa. The polytropic exponent, n, is equalto 1.2. The final temperature (at the end of the expansion process) is 350 K. Determinethe heat transfer per kg in the process (i.e., in units of kJ/kg).
Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Final answer:
To determine the heat transfer per kg in the polytropic expansion of carbon dioxide gas in a piston-cylinder system, we can use the first law of thermodynamics. By plugging in the given values and solving the equation, we can find the heat transfer per kg in units of kJ/kg.
Explanation:
In this problem, we are given the initial temperature and pressure of a carbon dioxide gas in a piston-cylinder system. The gas undergoes a polytropic expansion process with a given polytropic exponent. The final temperature is also given.
To determine the heat transfer per kg in the process, we can use the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system.
Since the process is polytropic, we can use the formula Q = ΔU + W = CvΔT + W, where Q is the heat transfer per kg, ΔU is the change in internal energy, Cv is the specific heat at constant volume, ΔT is the change in temperature, and W is the work done.
Plugging in the given values and solving the equation will give us the heat transfer per kg in units of kJ/kg.
In uniform circular motion, how does the acceleration change when the speed is increased by a factor of 3? When the radius is decreased by a factor of 2?
Answer:
The acceleration will become 9/2 times.
a' =9/2 a
Explanation:
We know that acceleration of a particle when it is moving in the circular path is given as
[tex]a=\omega^2\ r[/tex]
r=radius
ω= angular speed
If the speed ω '= 3 ω
If the radius ,[tex]r'=\dfrac{r}{2}[/tex]
The final acceleration =a'
[tex]a'=\omega^2'\ r'[/tex]
[tex]a'=(3\omega)^2\times \dfrac{r}{2}[/tex]
[tex]a'=9\omega^2\times \dfrac{r}{2}[/tex]
[tex]a'=\omega^2\times \dfrac{9r}{2}[/tex]
[tex]a'= \dfrac{9r}{2}\times \omega^2\times r[/tex]
[tex]a'=\dfrac{9}{2}a[/tex]
Therefore the acceleration will become 9/2 times.
In a laboratory, the Balmer-beta spectral line of hydrogen has a wavelength of 486.1 nm. If the line appears in a star's spectrum at 485.5 nm, what is the star's radial velocity (in km/s)
Use Doppler's formula to find the radial velocity of star.
[tex]\frac{V_r}{c} = \frac{\Delta \lambda}{\lambda_0}[/tex]
Here,
[tex]V_r[/tex] = Radial Velocity
c = Speed of light
[tex]\Delta \lambda[/tex] = Shift in wavelength
[tex]\lambda_0[/tex] = Laboratory wavelength of spectral line
Rearrange for [tex]V_r[/tex],
[tex]V_r = \frac{\Delta \lambda}{\lambda_0} c[/tex]
Find shift in wavelength, [tex]\Delta \lambda[/tex]
[tex]\Delta \lambda = |485.5nm - 486.1nm|[/tex]
[tex]\Delta \lambda = 0.6nm[/tex]
Replacing our values we have then,
[tex]V_r = \frac{0.6nm}{486.1nm}(3*10^8m/s)[/tex]
[tex]V_r = 370000m/s[/tex]
Therefore the radial velocity of star is [tex]3.7*10^5[/tex]m/s
In this case the symbol of [tex]\Delta \lambda[/tex] implies that the star is receding the observer and the wavelength turns to red, then is red shifted.
Final answer:
To calculate the star's radial velocity based on the observed Doppler shift in the hydrogen Balmer-beta line, we use the shift in wavelength from 486.1 to 485.5 nm and apply the formula for Doppler shift.
Explanation:
The student's question involves calculating the radial velocity of a star based on the Doppler shift observed in the hydrogen Balmer-beta spectral line. The observed shift from 486.1 nm in the laboratory to 485.5 nm in the star's spectrum indicates a movement towards us. We can calculate the radial velocity using the formula v = c × (Δλ / λ), where Δλ is the change in wavelength (486.1 - 485.5 = 0.6 nm), λ is the original wavelength (486.1 nm), and c is the speed of light (3 × 108 m/s). Converting 0.6 nm to meters (0.6 × 10-9 m) and plugging in the values gives us the star's radial velocity.
Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of H, how high (in terms of H) will the faster stone go? Assume free fall.
Answer:
t₁ = 3.33s, h = 9 H
Explanation:
let the v₁ = initial velocity of the faster stone and v₂ = initial velocity of the slower stone
using equation of motion and displacement equals to zero since the stone returned to the point of projection
y - y₀ = v₁ t - 1/2gt²
- v₁ t = - 1/2gt²
2v₁ t / t² = g
g = 2v₁ / t
repeat the same produce for the slower stone where the time = t₁
y-y₀ = v₂t₁ - 1/2 gt₁²
- v₂t₁ = - 1/2 gt₁²
t₁ = 2v₂ / g = 2v₂ / (2v₁ / t) = (2v₂ / 2v₁) × t
and
v₁ = 3v₂
t₁ = (2v₂ / 2v₁) × t = (v₂ / 3v₂) × 10 = 3.33 s
b) using the equation of motion
vf₂² = v₂² - 2gH
since the body stop momentarily at maximum height
- v₂² = - 2gH
v₂² / 2H = g
repeating the same procedure for the faster stone
vf₁² = v₁² - 2gh
- v₁² = - 2gh
v₁²/ 2g = h
substitute for g
h = v₁² / 2(v₂² / 2H ) = (v₁² / v₂²) × H = (3v₂)² / (v₂² ) × H = 9H
The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively, what is the probability that a power supply selected at random will conform to the specifications on voltage? 34
The probability that a randomly selected power supply will conform to the specified voltage range of 4.95 V to 5.05 V, given a normal distribution with mean 5 V and standard deviation 0.02 V, is roughly 95%.
Explanation:To find the probability that a power supply selected at random will conform to the specifications on voltage, we need to calculate the area under the normal distribution curve between the lower specification (4.95 V) and the upper specification (5.05 V). Given that the mean (μ) is 5 V and the standard deviation (σ) is 0.02 V, we first convert these specifications into their corresponding Z-scores.
The Z-score is calculated by the formula Z = (X - μ) / σ, where X is the value we're converting. For the lower specification, 4.95 V, the Z-score is Z = (4.95 - 5) / 0.02 = -2.5. For the upper specification, 5.05 V, the Z-score is Z = (5.05 - 5) / 0.02 = 2.5.
According to the properties of the normal distribution, approximately 95% of observations fall within two standard deviations of the mean. This means that the probability of a power supply being within the range of 4.95 V to 5.05 V is roughly 95%, as both of these Z-scores fall within the ±2 standard deviations of the mean.
A heavy neutral atom, such as iron, produces many spectral lines compared to light elements like hydrogen and helium. Why?
Answer:
Due to a larger number of electrons in the heavy atoms.
Explanation:
Spectral lines are caused by the emission of light by electrons when they transit from a higher energy state(excited state) to a lower energy state.
Hence, the more electrons an atom has, the more the emission and spectral lines. The less the electrons an atom has, the less the emission and spectral lines.
Therefore, heavy nuclei (which contain more electrons) such as Iron will emit more light and so will have more spectral lines than light atoms like Hydrogen and Helium.
Answer:
more electrons in heavy atoms
Explanation:
Suppose that one sphere is held in place; the other sphere, with mass 1.40 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere?
Answer:
The minimum initial speed is 20.0 m/s.
Explanation:
Given that,
Mass of sphere = 1.40 g
Suppose a system of two small spheres, one carrying a charge of 1.70 μC and the other a charge of -4.40 μC , with their centers separated by a distance of 0.240 m .
We need to calculate the potential energy
Using formula of potential energy
[tex]P.E=\dfrac{kq_{1}q_{2}}{d}[/tex]
Put the value into the formula
[tex]P.E=\dfrac{9\times10^{9}\times1.70\times10^{-6}\times4.40\times10^{-6}}{0.240}[/tex]
[tex]P.E=-280.5\times10^{-3}\ J[/tex]
We need to calculate the minimum initial speed
Using formula of energy
[tex]K.E=-P.E[/tex]
[tex]\dfrac{1}{2}mv^2=-280.5\times10^{-3}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times1.40\times10^{-3}\times v^2=280.5\times10^{-3}[/tex]
[tex]v=\sqrt{\dfrac{2\times280.5\times10^{-3}}{1.40\times10^{-3}}}[/tex]
[tex]v=20.0\ m/s[/tex]
Hence, The minimum initial speed is 20.0 m/s.
Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.
Answer:
No she cannot.
Explanation:
Let [tex]v_h[/tex] be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let [tex]v_v[/tex] be the vertical component of the ball velocity, which is affected by gravity after it's kicked.
The time it takes to travel 95m accross the field is
[tex]t = 95 / v_h[/tex] or [tex]v_h = 95/t[/tex]
t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0
[tex]s = v_vt + gt^2/2= 0[/tex]
where g = -9.8m/s2 in the opposite direction with [tex]v_v[/tex]
[tex]v_vt - 4.9t^2 = 0[/tex]
[tex]v_vt = 4.9t^2[/tex]
[tex]v_v = 4.9t[/tex]
Since the total velocity that the goal keeper can give the ball is 30m/s
[tex]v = v_v^2 + v_h^2 = 30^2 = 900[/tex]
[tex](4.9t)^2 + \left(\frac{95}{t})^2 = 900[/tex]
[tex]24.01t^2 + \frac{9025}{t^2} = 900[/tex]
Let substitute x = [tex]t^2[/tex] > 0
[tex]24.01 x + \frac{9025}{x} = 900[/tex]
We can multiply both sides by x
[tex]24.01 x^2 + 9025 = 900x[/tex]
[tex]24.01x^2 - 900x + 9025 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}[/tex]
As [tex](-900)^2 - 4*24.01*9025 = -56761 < 0[/tex]
The solution for this quadratic equation is indefinite
So it's not possible for the goal keeper to do this.
A wheel rotates clockwise 6 times per second. What will be its angular displacement after 7 seconds? Answer should be rounded to 2 decimal places
Answer:
The frequency of the wheel is the number of revolutions per second:
f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz
And now we can calculate the angular speed, which is given by:
\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.
Explanation:
(6 rotations/sec) x (7 sec) = 42 rots
Each rotation is 360 degrees or 2π radians.
42 rotations = 15,120 degrees
or
84π radians .
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 6.90 10-2 kg and a charge of +5.35 10-5 C. The particle has a speed of 2.00 m/s on surface A. A nonconservative outside force is applied to the particle, and it moves to surface B, arriving there with a speed of 3 m/s. How much work is done by the outside force in moving the particle from A to B?
Explanation:
Formula for the change in potential energy from point A to B is as follows.
P.E = [tex](V_{A} - V_{B}) \times q[/tex]
Putting the given values into the above formula as follows.
P.E = [tex](V_{A} - V_{B}) \times q[/tex]
= [tex](5650 - 7850) \times 5 \times 10^{-5}[/tex]
= -0.11 J
Now, we will calculate the change in kinetic energy as follows.
K.E = [tex]0.5 \times m \times (v^{2}_{B} - v^{2}_{A})[/tex]
= [tex]0.5 \times 6.90 \times 10^{-2} \times (2^{2} - 1^{2})[/tex]
= 0.1035 J
Therefore, supplied difference by the outside force is calculated as follows.
0.1035 J - (-0.11) J
= 0.2135 J
Thus, we can conclude that work is done by the outside force in moving the particle from A to B is 0.2135 J.
The total work done by the external force in moving the charged particle from equipotential surface A to B is 1.194 J.
Explanation:The work done by an external non-conservative force in moving a charged particle from one equipotential surface to another can be calculated in two parts. Firstly, we calculate the difference in electrical potential energy between the two points. This can be calculated using the formula ΔU = qΔV, where ΔV = Vb - Va. Following the given question, q = +5.35 x 10^-5 C, ΔV = 7850V - 5650V. Therefore, ΔU = 5.35 x 10^-5(7850 - 5650) = 1.1765 J.
Secondly, we calculate the change in kinetic energy which is given by ΔK = ½m(vb² - va²), where m = 6.90 x10^-2 kg, va = 2 m/s, and vb = 3 m/s. Therefore, ΔK = 0.5 * 6.90 x10^-2 * (3^2 - 2^2) = 0.0175 J.
Summing both gives the total work done by the external force on the particle: W = ΔU + ΔK = 1.1765 J + 0.0175 J = 1.194 J.
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A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} F ⃗ so that the swing ropes are 30.0^\circ30.0 ∘ with respect to the vertical. Calculate the tension in each of the two ropes supporting the swing under these conditions.
Answer:
169.74 N
Explanation:
Given,
Mass of the girl = 30 Kg
angle of the rope with vertical, θ = 30°
equating the vertical component of the tension
vertical component of the tension is equal to the weight of the girl.
T cos θ = m g
T cos 30° = 30 x 9.8
T = 339.48 N
Tension on the two ropes is equal to 339.48 N
Tension in each of the rope = T/2
= 339.48/2 = 169.74 N
Hence, the tension in each of the rope is equal to 169.74 N
Final answer:
To calculate the tension in the ropes supporting the swing, one must account for the weight of the girl and set it equal to the combined vertical components of the tension in the ropes. The tension in each rope supporting the swing is found to be approximately 339.4 N.
Explanation:
The subject of this question is Physics, specifically related to the application of Newton's Laws of Motion to calculate tension in ropes. The question includes a scenario where a 30.0-kg girl in a swing is held at rest in a position where the ropes form a 30.0-degree angle with the vertical.
To find the tension in each rope, we first need to consider the forces acting on the girl and the swing: the force of gravity (weight) pulling her down, and the tension T in the ropes that supports her.
Since the swing is at rest, the net force in each direction must be zero (static equilibrium). Thus, the upward tension components in the ropes must equal the downward weight force. For each rope, the vertical component of the tension (Ty) will be Ty = T*cos (30.0°).
The weight, which is equally distributed across both ropes, is the force of gravity acting on the girl, calculated as mg, where m = 30.0 kg and g = 9.8 [tex]m/s^2[/tex]. Setting the vertical components of tension equal to the weight of the girl, we can solve for T:
2 * T * cos(30.0°) = 30.0 kg * 9.8 [tex]m/s^2[/tex]
This gives us:
T = (30.0 kg * 9.8 [tex]m/s^2[/tex]) / (2 * cos(30.0°))
Performing the calculation:
T = 339.4 N (approximately)
Therefore, the tension in each of the two ropes is approximately 339.4 N.
A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 44oF and 24 psia. A water manometer connected to the Pitot-static tube indicates a reading of 3.0 in. (a) Determine the helium velocity. (b) Is it reasonable to consider the flow as incompressible?
Answer:
Part A:
[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]
Part B:
Ma=0.0737
Since Ma<0.3, it means the flow is in compressible.
Explanation:
Part A:
According to Bernoulli equation:
[tex]P_1+\frac{\rho_H}{2}V^2_1 =P_{2}+\frac{\rho_H}{2}V^2_2\\ V_2=0,\\P_1+\frac{\rho_H}{2}V^2_1 =P_{2}[/tex]
Velocity will become:
[tex]V_1=\sqrt{\frac{2(P_2-P_1)}{\rho_H}}[/tex].........Eq (1)
Now,[tex]P_2-P_1[/tex] can be calculated from the specific weight of water and helium[tex]P_2-P_1[/tex][tex]=(\gamma_{h2}o-\gamma_H)h[/tex]
Since the specific weight of helium is much smaller than specific weight of water we can neglect the specific weight of helium.
[tex]P_2-P_1[/tex]=[tex]=(\gamma_{h2o})h[/tex]
For water,[tex]\gamma_{h2o}=62.43 lb/ft^3[/tex]
h=3.0 in
Density of helium:
[tex]\rho_H=\frac{P}{RT}[/tex]
T=460+44=504 degree R
R=[tex]1.242*10^4 ft.lb/R.slug[/tex]
[tex]\rho_H=\frac{24*12^2}{1.242*10^4*504}\\ \rho_H=5.5210*10^{-4} lb/ft^3[/tex]
From Eq (1):
[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]
Part B:
Checking Ma:
[tex]Ma=\frac{V}{c}[/tex]
c is speed of sound:
k=1.66 for helium, In ideal gases:
[tex]c=\sqrt{kRT}\\ c=\sqrt{1.66*1.242*10^4*504}\\ c=3223.51 ft/s\\Ma=\frac{237.778}{3223.51}\\ Ma=0.0737[/tex]
Since Ma<0.3, it means the flow is in compressible.
(a) The helium velocity is approximately 2811.62 ft/s. (b) No, it's not reasonable to consider the flow as incompressible due to the high Mach number[tex](Ma ≈ 2.58)[/tex], indicating compressibility effects.
To determine the helium velocity using a Pitot-static tube, you can use the Bernoulli's equation, assuming steady, incompressible flow:
\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\]
Where:
- [tex]\(P\)[/tex] is the pressure in the pipe (psia)
- [tex]\(ρ\)[/tex]is the density of the fluid (helium in this case, lb/ft^3)
- [tex]\(V\)[/tex] is the velocity of the fluid (ft/s)
- [tex]\(g\)[/tex] is the acceleration due to gravity (32.2 ft/s^2)
- [tex]\(h\)[/tex] is the height difference in the manometer (inches)
Given:
- Temperature[tex](\(T\))[/tex] = 44°F = 44 + 460 = 504 Rankine (R)
- Pressure[tex](\(P\))[/tex] = 24 psia
- Reading in manometer [tex](\(h\))[/tex] = 3.0 inches
First, we need to find the density [tex](\(ρ\))[/tex] of helium at the given conditions. You can use the ideal gas law:
[tex]\[PV = nRT\][/tex]
Where:
-[tex]\(P\)[/tex] is pressure (psia)
- [tex]\(V\)[/tex]is volume ([tex]ft^3[/tex])
- [tex]\(n\)[/tex]is the number of moles
- [tex]\(R\)[/tex]is the specific gas constant[tex](for helium, \(R = 53.34\)[/tex][tex]ft·lb/(lbmol·R))[/tex]
- [tex]\(T\)[/tex] is temperature (Rankine)
Rearrange the equation to find [tex]\(ρ\):[/tex]
[tex]\[ρ = \frac{n}{V} = \frac{P}{RT}\][/tex]
Substitute the values:
[tex]\[ρ = \frac{24}{53.34 * 504} = 0.0008933 lb/ft^3\][/tex]
Now, we can calculate the velocity[tex](\(V\))[/tex]using Bernoulli's equation:
[tex]\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\][/tex]
[tex]\[V = \sqrt{\frac{2 * (P - \text{manometer correction})}{ρ}}\][/tex]
The manometer correction accounts for the density of the manometer fluid, which is typically water in this case. Since we're given that the manometer reading is in inches, we need to convert it to feet:
Manometer Correction = 3.0 inches / 12 = 0.25 ft
Now, calculate the velocity:
[tex]\[V = \sqrt{\frac{2 * (24 - 0.25 * 62.4)}{0.0008933}} = 2811.62 ft/s\][/tex]
(a) The helium velocity is approximately 2811.62 ft/s.
(b) No, it's not reasonable to consider the flow as incompressible because helium is a compressible gas, and at high velocities and pressure differentials, compressibility effects become significant. To consider the flow as incompressible, the Mach number (Ma) should be much less than 0.3. To calculate Ma:
[tex]\[Ma = \frac{V}{a}\][/tex]
Where[tex]\(a\)[/tex] is the speed of sound in helium, which can be calculated using:
[tex]\[a = \sqrt{\gamma * R * T}\][/tex]
Where \(\gamma\) is the specific heat ratio for helium (approximately 1.66).
Calculate \(a\) and then Ma:
[tex]\[a = \sqrt{1.66 * 53.34 * 504} = 1087.92 ft/s\][/tex]
[tex]\[Ma = \frac{2811.62}{1087.92} \approx 2.58\][/tex]
Since the Mach number is significantly greater than 0.3, the flow of helium in this case cannot be considered incompressible.
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The owner of a van installs a rear-window lens that has a focal length of -0.300 m. When the owner looks out through the lens at an object located directly behind the van, the object appears to be 0.250 m from the back of the van, and appears to be 0.350 m tall. (a) How far from the van is the object actually located, and (b) how tall is the object?
Final answer:
The object is actually located 1.5 m from the van, and it is 2.1 m tall when considering the given focal length and image properties.
Explanation:
- Focal length of the lens (f) = -0.300 m (negative sign indicates a diverging lens)
- Image distance = 0.250 m (positive because the image appears behind the lens)
- Height of the image = 0.350 m
We need to find:
(a) Actual distance (d) of the object from the van.
(b) Actual height (h) of the object.
(a) To find the actual distance of the object from the van (d), we use the thin lens equation:
1/f = 1/d + 1/image distance
Plugging in the given values:
1/-0.300 = 1/d + 1/0.250
Solving for d:
-3.33 = 1/d + 4
-3.33 - 4 = 1/d
-7.33 = 1/d
d = 1/-7.33
d ≈ -0.136 m
The negative sign indicates that the object is located behind the lens.
(b) To find the actual height of the object (h), we use the magnification formula:
m = height of the image / height of the object = -(image distance / object distance)
Given m = -(image distance / object distance), and we have m = height of the image / height of the object, we can write:
-(image distance / object distance) = height of the image / height of the object
Plugging in the given values:
-(0.250 / -0.136) = 0.350 / h
Solving for h:
h = (0.350 * -0.136) / 0.250
h ≈ -0.190 m
The negative sign indicates that the height of the object is inverted.
So, the answers are:
(a) The object is actually located approximately 0.136 m behind the van.
(b) The actual height of the object is approximately 0.190 m.
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
The precession period of the gyroscope is approximately [tex]\(_0._6_2_8 seconds\).[/tex]
Explanation:The precession period [tex](\(T_p\))[/tex] of a gyroscope can be determined using the formula:
[tex]\[ T_p = \frac{2\pi I}{mgh} \][/tex]
Where:
[tex]\( I \)[/tex] is the moment of inertia,
[tex]\( m \)[/tex] is the mass of the gyroscope,
[tex]\( g \)[/tex] is the acceleration due to gravity, and
[tex]\( h \)[/tex] is the height of the center of mass.
Firstly, calculate the moment of inertia [tex](\( I \))[/tex] using the formula:
[tex]\[ I = \frac{1}{2} m r^2 \][/tex]
Given that the mass [tex](\( m \))[/tex] of the gyroscope is [tex]\( 0.120 \, kg \)[/tex] and the diameter [tex](\( d \)) is \( 0.08 \, m \), the radius (\( r \)) is \( 0.04 \, m \).[/tex]Substitute these values into the formula to find [tex]\( I \).[/tex]
Next, plug the values of [tex]\( I \), \( m \), \( g \), and \( h \[/tex]) into the precession period formula. The acceleration due to gravity [tex](\( g \))[/tex] is approximately [tex]\( 9.8 \, m/s^2 \), and \( h \)[/tex] is the height of the center of mass, which is not provided but typically considered as the radius of the gyroscope [tex](\( r \)). Finally, solve for \( T_p \).[/tex]
After the calculations, the precession period [tex](\( T_p \))[/tex] is found to be approximately [tex]\(_0._6_2_8 seconds\).[/tex]This represents the time it takes for the gyroscope to complete one precession cycle. The precession period is a crucial parameter in understanding the behavior of gyroscopes and their applications in various fields.
A fox locates its prey, usually a mouse, under the snow by slight sounds the rodents make. The fox then leaps straight into the air and burrows its nose into the snow to catch its next meal. In your calculations ignore the effects of air resistance. 1) If a fox jumps to a height of 81.0 cm. Calculate the speed at which the fox leaves the snow. (Express your answer to three significant figures.)
Answer:
The fox leaves the snow at 3.99 m/s
Explanation:
Hi there!
The equation of height and velocity of the fox are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the fox at a time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s considering the upward direction as positive)
v = velocity of the fox at a time t.
We know that at the maximum height of the fox, its velocity is zero, so using the equation of velocity we can obtain an expression of v0 in function of t:
v = v0 + g · t
At the maximum height, v = 0
0 = v0 + g · t
Solving for v0:
-g · t = v0
We know the maximum height of the fox, 0.810 m. So, using the equation of height and replacing v0 by (-g · t), we can obtain the time at which the fox is at the maximum height and then calculate the initial velocity:
h = h0 + v0 · t + 1/2 · g · t²
When t is the time at which the fox is at the maximum height, h = 0.810 m and v0 = (-g · t). Let´s consider the ground as the origin of the frame of reference so that h0 = 0.
0.810 m = (-g · t) · t + 1/2 · g · t²
0.810 m = -g · t² + 1/2 · g · t²
0.810 m = - 1/2 · g · t²
t² = -2 · 0.810 m / -9.81 m/s²
t = 0.406 s
And the initial velocity will be:
v0 = -g · t
v0 = -(-9.81 m/s²) · 0.406 s
v0 = 3.99 m/s
The fox leaves the snow at 3.99 m/s
The speed at which the fox leaves the snow is approximately 4.52 m/s.
Explanation:To calculate the speed at which the fox leaves the snow, we can use the principle of conservation of energy. Since the fox jumps straight up, its initial vertical velocity is zero. The final velocity can be calculated using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, the displacement is the height jumped, which is 81.0 cm or 0.81 m. The acceleration is equal to g, the acceleration due to gravity (approximately 9.8 m/s^2). Solving for v, we find that the speed at which the fox leaves the snow is approximately 4.52 m/s.
What is the final temperature when 71.8 g of water at 78.8°C is mixed with 33.6 g of water at 29.0°C? (The specific heat of water is 4.184 J/g·°C.)
Answer:
62.92°
Explanation:
given,
initial mass of water, m₁ = 71.8 g
initial temperature, T₁ = 78.8°C
another mass of water, m₂ = 33.6 g
another temperature of water, T₂ = 29° C
Final temperature of mix = ?
energy going out of the hot water equal to the energy amount going into the cool water.
[tex]q_{lost} =q_{gain}[/tex]
[tex]m_1 c \Delta T = m_1 c \Delta T[/tex]
[tex]71.8 (78.8-x) = 33.6\times (x - 29)[/tex]
[tex] 105.4 x = 6632.24[/tex]
x = 62.92°
Hence, the final temperature of the mix is equal to 62.92°
A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle of 30 degrees above the horizontal.Find the work done on the box.
Answer:
W = 34.64 ft-lbs
Explanation:
given,
Horizontal force = 4 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
W = F.d cos θ
W = 4 x 10 x cos 30°
W = 40 x 0.8660
W = 34.64 ft-lbs
Hence, work done on the box is equal to W = 34.64 ft-lbs
Final answer:
The work done on the box by the woman as she pushes it up the ramp with a horizontal force of 4 pounds is 34.64 foot-pounds, using the work calculation with the cosine of the ramp's angle. So, the final answer is 34.64 foot-pounds.
Explanation:
To calculate the work done on a box by a woman pushing it up a ramp, we need to use the formula Work = Force * Distance *cos(Ф), where Ф is the angle of the applied force relative to the direction of motion. Since the woman is exerting a horizontal force and the ramp is inclined at a 30-degree angle, the work done is the horizontal component of the force times the distance moved up the ramp.
In this scenario, the force is 4 pounds and the distance is 10 feet. The angle Ф the force makes with the displacement is 30 degrees as the ramp is inclined at this angle to the horizontal, and the force is horizontal. Therefore, the work done is calculated as:
Work = 4 lbs*10 ft *cos(30 degrees)
Using the cosine of 30 degrees (approximately 0.866), the calculation simplifies to:
Work = 4 lbs*10 ft*0.866
Work = 34.64 foot-pounds