Answer:
If the radius of the disk is much greater than the point where the electric field is calculated, then the field of the disk matches the field of the infinite sheet.
Explanation:
First, we have to calculate the electric field of the disk.
We should choose an infinitesimal area, 'da', on the disk and calculate the E-field of this small portion, 'dE'. Then we will integrate dE over the entire disk using cylindrical coordinates.
According to the cylindrical coordinates: da = rdrdθ
The small portion is chosen at a distance r from the axis. Let's find the dE at a point on the axis and a distance z from the center of the disk.
[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2 + z^2}[/tex]
Here dQ can be found by the following relation: The charge density of the disk is equal to the total charge divided by the total area of the disk. The small portion of the disk will have the same charge density, therefore:
[tex]\frac{Q}{\pi R^2} = \frac{dQ}{da}\\dQ = \frac{Qda}{\pi R^2}[/tex]
Furthermore, we need to separate the vertical and horizontal components of dE, because it is a vector and cannot be integrated without separating the components. By symmetry, the horizontal components of dE will cancel out each other, leaving only the vertical components in the z-direction.
[tex]dE_z = dE\sin(\alpha) = dE \frac{z}{\sqrt{z^2+r^2}}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qzda}{\pi R^2(z^2+r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}[/tex]
We have to use a double integral over the radius and the angle to find the total electric field due to a uniformly charged disk:
[tex]E = \int \int dE = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 {\int\limits^R_0 {\frac{1}{(z^2+r^2)^{3/2}}} \, rdr} \, d\theta\\E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[1 - \frac{1}{\sqrt{(R^2/z^2) + 1}}][/tex]
If the radius of the disk is much greater than the point z, R >> z, than the term in the denominator becomes very large, and the fraction becomes zero. In that case electric field becomes
[tex]E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[/tex]
This is equal to the electric field of an infinite sheet.
As a result, the condition for the field of a disk to be equal to that of a infinite sheet is R >> z.
A spaceship takes off vertically from rest with an acceleration of 30.0 m/s 2 . What magnitude of force F is exerted on a 53.0 kg astronaut during takeoff?
Answer:
1590 N.
Explanation:
Force: This can be defined as the product of mass and the acceleration of a body. The S.I unit of Force is Newton (N).
The formula of force is given as
F = ma ........................ Equation 1
Where F = Force, m = mass of the astronaut, a = takeoff acceleration if the astronaut.
Given: m = 53.0 kg, a = 30 m/s²
Substitute into equation 1
F = 53(30)
F = 1590 N.
Hence the force exerted on the astronaut = 1590 N.
To calculate force, use the equation F=ma. Given an astronaut's mass of 53.0 kg and acceleration of 30.0 m/s² during a spaceship's takeoff, the exerted force totals around 1071 Newtons, subtracting the force of gravity.
Explanation:To calculate the force exerted on the astronaut, you would use the equation F=ma, where F is force, m is mass and a is acceleration. Given that the mass (m) is 53.0 kg and the acceleration (a) is 30.0 m/s², the force (F) can be calculated as follows: F = ma = (53.0 kg)(30.0 m/s²) = 1590 Newtons.
This force is a result of combining gravity and the spaceship's upward propulsion (i.e. the astronaut's weight and the force of the spaceship accelerating). The force of gravity can be calculated simply by Fgravity = mg = (53.0 kg)(9.8 m/s²) = 519.4 Newtons. Therefore, the net force exerted by the spaceship on the astronaut during takeoff is F - Fgravity = 1590N - 519.4N = 1070.6 N, rounded off to 1071 N.
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A fixed system of charges exerts a force of magnitude 9.0 N on a 3.0 C charge. The 3.0 C charge is replaced with a 10.0 C charge. What is the exact magnitude of the force (in N) exerted by the system of charges on the 10.0 C charge?
Answer:
30 N
Explanation:
We are given that
Force,F =9 N
Charge,q=3 C
We have to find the exact magnitude of the force(in N) exerted by the system of charges on the 10.0 C.
We know that
[tex]E=\frac{F}{q}[/tex]
Using the formula
[tex]E=\frac{9}{3}=3 N/C[/tex]
Now, charge=10 C
The electric field remains same then
[tex]F=3\times 10=30 N[/tex]
Hence, the exact magnitude of the force exerted by the system of charges on the 10 C=30 N
The exact magnitude of the force exerted by the system of charges on the 10.0 C charge is 29.97 N.
Explanation:The force exerted by a fixed system of charges on a charge is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, we are given that the force exerted on a 3.0 C charge is 9.0 N. When the 3.0 C charge is replaced with a 10.0 C charge, the force can be calculated using the proportionality of the charges. Since the charge is increased by a factor of (10.0 C)/(3.0 C) = 3.33, the force will also increase by the same factor. Therefore, the exact magnitude of the force exerted by the system of charges on the 10.0 C charge is 9.0 N * 3.33 = 29.97 N.
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If a beam of electromagnetic radiation has an intensity of 120 W/m2, what is the maximum value of the electric field?
Answer:
Electric field, E = 300.65 N/C
Explanation:
Given that,
Intensity of a beam of electromagnetic radiation, [tex]I=120\ W/m^2[/tex]
We need to find the maximum value of the electric field. The intensity of electromagnetic wave in terms of electric field is given by :
[tex]I=\dfrac{1}{2}\epsilon_oE^2c[/tex]
c is the speed of light
[tex]E=\sqrt{\dfrac{2I}{\epsilon_o c}}[/tex]
[tex]E=\sqrt{\dfrac{2\times 120}{8.85\times 10^{-12}\times 3\times 10^8}}[/tex]
E = 300.65 N/C
So, the maximum value of the electric field is 300.65 N/C. Hence, this is the required solution.
The maximum value of the electric field for the given beam of electromagnetic radiation.
the electric field of an electromagnetic radiation can be calculated by,
[tex]\bold {I = \dfrac 12 \epsilon_0 E^2C}[/tex]
[tex]\bold {E = \sqrt {\dfrac {2I}{\epsilon_0 C}}}[/tex]
Where,
I - intensity of the beam = 120 W/m2
C- speed of light = [tex]\bold { 3x10^8\ m/s}}[/tex]
E - electric field
Put the value of the formula
[tex]\bold {E = \sqrt {\dfrac {2\times 20 }{8.85x10^{-12} \times 3x10^8}}}\\\\\bold {E = 300.65\ N/C}[/tex]
Therefore, the maximum value of the electric field for the given beam of electromagnetic radiation.
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A sample of gas occupies a volume of 57.8 mL 57.8 mL . As it expands, it does 110.8 J 110.8 J of work on its surroundings at a constant pressure of 783 Torr 783 Torr . What is the final volume of the gas?
Answer: The final volume of the gas is 1.12 L
Explanation:
W = work done on or by the system
w = work done by the system = [tex]P\Delta V[/tex]
P = pressure = 783 torr = 1.03 atm (1atm=760 torr)
w= 110.8 J = 1.094 J (1Latm=101.3J)
[tex]V_1[/tex] = initial volume = 57.8 ml = 0.0578L (1L=1000ml)
[tex]V_2[/tex] = final volume = ?
[tex]1.094Latm=1.03atm\times (V_2-0.0578)L[/tex]
[tex](V_2-0.0578)=1.06[/tex]
[tex]V_2=1.12L[/tex]
The final volume of the gas is 1.12 L
A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The time units are seconds and the distance units are feet. The distance between the location of the pitcher and homeplate (where the batter stands) is 60.5 feet. Give EXACT answers, unless instructed otherwise. (a) Calculate the horizontal velocity of the baseball at time t; this is the function x'(t)= 142 Correct: Your answer is correct. ft/sec. (b) What is the horizontal velocity of the baseball when it passes over homeplate? 142 Correct: Your answer is correct. ft/sec (c) What is the vertical velocity of the baseball at time t; this is the function y'(t)= $$−32t+5 Correct: Your answer is correct. -32t +5 ft/sec. (d) Recall that the speed of the baseball at time t is s(t)=√ [x '(t)]2 + [y ' (t)]2 ft/sec. What is the speed of the baseball (in mph) when it passes over homeplate? $$1 Incorrect: Your answer is incorrect. (sqrt(142*142 +((-1936/142) + 5)**2)*(360/528)) mph. (e) At what time does the baseball hit the ground, assuming the batter and catcher miss the ball? $$1.1 Incorrect: Your answer is incorrect. (5+sqrt(5**2 + 320))/32 sec. (f) What is the magnitude of the angle at which the baseball hits the ground? 0.12 Incorrect: Your answer is incorrect. rad. (This is the absolute value of the angle between the tangential line to the path of the ball and the ground. Give your answer in radians to three decima
The question involves calculations using the principles of projectile motion and 2D kinematics to understand the trajectory of a baseball pitch. Values include horizontal and vertical velocity changes over time, the exact speed of the baseball and the angle at which baseball hits the ground.
Explanation:The question involves the physics of projectile motion, specifically applied to the trajectory of a pitch in baseball. The parameters of this problem can be solved using the kinematic equations and principles that govern 2D motion.
(1) To solve part (a), one must understand that for such problems involving gravity, the x (horizontal) component of the object's velocity remains constant throughout the motion, in this case 142 ft/sec. This fact is obtained from the equation x(t) = v_xt, where v_x is the constant x-component (horizontal) of velocity.
(2) For part (b), similarly, the x-component of the velocity remains same at any point in trajectory, including when it passes over home plate, so it is again 142 ft/sec.
(3) For part (c), the derivative of y(t) -16t^2 + 5t + 5 needs to be found to get the y-component (vertical) of velocity which is -32t + 5. This is because as the object moves under the effect of gravity, vertical velocity changes with time.
(4) For part (d), the magnitude of the speed at any time can be found by taking the square root of the sum of squares of x and y-components of velocity. (5) Part (e), seeks the time when the baseball hits the ground. The equation of vertical motion -16t^2 + 5t + 5 = 0 should be used, as the height of the object from ground is 0 when it hits the ground. Solving for t would give that time.
End of the problem involves understanding some trigonometric relationships to calculate the angle (in radians).
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A uniformly accelerating rocket is found to have a velocity of 11.0 m/s when its height is 4.00 m above the ground, and 1.90 s later the rocket is at a height of 56.0 m. What is the magnitude of its acceleration?
Answer:
17.23 m/s²
Explanation:
Applying the equation of motion,
Δs = ut + 1/2at²..................... Equation 1
Where Δs = change in height of the rocket, u = initial velocity of the rocket, a = acceleration of the rocket, t = time
making a the subject of the equation,
a = 2(Δs-ut)/t²..................... Equation 2
Given: Δs = (56-4) m = 52 m, u = 11.0 m/s, t = 1.90 s.
Substitute into equation 2
a = 2[52-(11×1.9)]/1.9²
a = 2(52-20.9)/1.9²
a = 2(31.1)/3.61
a = 62.2/3.61
a = 17.23 m/s².
Thus the acceleration = 17.23 m/s²
Final answer:
To find the magnitude of acceleration, we used the kinematic equation s = ut + ½at². We determined the displacement due to the initial velocity and the time interval, and then solved for acceleration to find that the rocket's magnitude of acceleration is 17.2 m/s².
Explanation:
To calculate the magnitude of its acceleration, we can use the kinematic equations for uniformly accelerated motion along with the information provided about the rocket's velocity and positions at different times. We have two positions, 4.00 m and 56.0 m, and a time interval of 1.90 s. The rocket's velocity at the lower height is 11.0 m/s.
Step 1: Use the second kinematic equation
We will use the kinematic equation:
s = ut + ½at², where 's' is the displacement, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.
Step 2: Set up the equation with the known values
Let's calculate the displacement (Δs): Δs = 56.0 m - 4.00 m = 52.0 m
Now put the known values into the equation: 52.0 m = (11.0 m/s)(1.90 s) + ½a(1.90 s)²
Step 3: Solve for acceleration 'a'
First, we calculate the distance traveled due to the initial velocity: (11.0 m/s)(1.90 s) = 20.9 m
Subtract this value from the total displacement to find the displacement caused by acceleration: 52.0 m - 20.9 m = 31.1 m
Rewrite the equation with this new information: 31.1 m = ½a(1.90 s)²
Now solve for 'a': a = (2 × 31.1 m) / (1.90 s)² = 17.2 m/s²
The magnitude of acceleration of the rocket is therefore 17.2 m/s².
A bullet is fired horizontally from the top of an ocean-based drill site. If the air resistance is negligible and the vertical distance to the water is 35.0 m, what additional information is needed to determine the time required for the bullet to hit the water? (g = 9.81 m/s2)
Answer:
no additional information required.
Explanation:
given,
vertical height, h = 35 m
acceleration due to gravity, g = 9.8 m/s²
time taken by the bullet to fall in the water = ?
initial vertical velocity of the bullet = 0 m/s
using equation of motion for the time calculation
[tex]s = ut + \dfrac{1}{2}gt^2[/tex]
[tex]h= 0\times t + \dfrac{1}{2}\times g \times t^2[/tex]
[tex]t = \sqrt{\dfrac{2h}{g}}[/tex]
[tex]t= \sqrt{\dfrac{2\times 35}{9.8}}[/tex]
t = 2.67 s
The time taken by the bullet to fall in the ocean is equal to 2.67 s.
Hence, there is no additional information required.
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true anomaly of 150 deg. If the speed of the meteoroid at that time is 2.23 km/s, calculate: (a) the eccentricity of the trajectory, (b) the altitude at closest approach, (c) the speed at the closest approach, (d) the aiming radius and turn angle, and (e) the C3 of the meteoroid. Sketch the orbit.
a: The eccentricity is 1.086.
b: The altitude at closest approach is 5088 km
c: The velocity at perigee is 8.516 km/s
d: The turn angle is 134.08 while the aiming radius is 5641.28 km
Part a
Specific energy is given by
[tex]\epsilon=(v^2)/(2)-(\mu)/(r)[/tex]
Here
ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km
[tex]\epsilon=(v^2)/(2)-(\mu)/(r)\n\epsilon=(2.2^2)/(2)-(398600)/(402,000)\n\epsilon=1.495 km^2/s^2[/tex]
Value of specific energy is also given as
[tex]\epsilon=(\mu)/(2a)\na=(\mu)/(2\epsilon)\na=(398600)/(2* 1.495)\na=13319 km[/tex]
Orbit formula is given as
[tex]r=a((e^2-1)/(1+ecos \theta))\nae^2-recos\theta-(a+r)=0[/tex]
Putting values in this equation and solving for e via the quadratic formula gives
[tex]ae^2-recos\theta-(a+r)=0\n(133319)e^2-(402000)(cos 150) e-(133319+402000)[/tex]
[tex]=0\n133319e^2+348142.21 e-535319=0\n\ne[/tex]
[tex]=(-348142.21 \pm \sqrt{(348142.21^2-4(133319)(535319)))}/(2 (133319))\n\ne[/tex]
=1.086 , or , -3.69
As the value of eccentricity cannot be negative so the eccentricity is 1.086.
Part b
The radius of trajectory at perigee is given as
[tex]r_p=a(e-1)\n[/tex]
Substituting values gives
[tex]r_p=133319 (1.086-1)\nr_p=11465.4 km[/tex]
Now for estimation of altitude z above earth is given as
[tex]z=r_p-R_E\nz=11465.4-6378\nz=5087.434\nz\approx 5088 km[/tex]
So the altitude at closest approach is 5088 km
Part c
radius of perigee is also given as
[tex]r_p=(h^2)/(\mu)(1)/(1+e)[/tex]
Rearranging this equation gives
[tex]h=√(r_p\mu(1+e))\nh=√(11465.4 * 3986000 * (1+1.086))\nh=97638.489 km^2/s[/tex]
Now the velocity at perigee is given as
[tex]v_p=(h)/(r_p)\nv_p=(97638.489)/(11465.4)\nv_p=8.516 km/s\n[/tex]
So the velocity at perigee is 8.516 km/s
Part d
Turn angle is given as
[tex]\delta =2 sin^{(-1)} ((1)/(e))[/tex]
Substituting value in the equation gives
[tex]\delta =2 sin^{(-1)} ((1)/(e))\n\delta =2 sin^{(-1)} ((1)/(1.086))\n\delta =134.08[/tex]
Aiming radius is given as
[tex]\Delta =a \sqrt{(e^2-1)[/tex]
Substituting value in the equation gives
[tex]\Delta =a \sqrt{(e^2-1)\n\Delta} =13319 \sqrt{(1.086^2-1)\n\Delta}=5641.28 km[/tex]
So the turn angle is 134.08 while the aiming radius is 5641.28 km
Therefore, a: The eccentricity is 1.086.
b: The altitude at closest approach is 5088 km
c: The velocity at perigee is 8.516 km/s
d: The turn angle is 134.08 while the aiming radius is 5641.28 km
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet, she throws a rock straight upward with an initial velocity of 20 m/s. If the astronaut were instead on Earth, and threw a ball in the same way while standing on a 50.0 m high cliff, what would be the time difference (in s) for the rock to hit the ground below the cliff in each case
Using kinematics equations, we can calculate the time for a rock to fall from the cliff on Earth. Without the value of acceleration due to gravity on the extrasolar planet, we can't quantify the time difference. A difference would exist if the gravitational accelerations of the two planets differ.
Explanation:This question pertains to kinematics and physical laws regarding free fall. On earth, when the astronaut throws a rock straight up with a certain velocity, initially, gravity will slow down the rock and finally make it fall back toward the ground. This action is governed by the equation of motion: h = vit + 1/2gt² where h is the height, vi the initial velocity, t the time and g the acceleration due to gravity on earth (-9.81 m/s²).
In the extrasolar planet scenario, considering it is not mentioned, we could assume that the acceleration due to gravity might be similar to earth's. However, without specific details about the gravity, we can still calculate the time it takes for the rock to land from the 50m high cliff using the motion equation for a free-falling body. We can conclude that if the gravity in the two settings is different, then there would be a time difference. We can’t quantify the time difference without knowing the value of the gravity on the extrasolar planet.
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A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car of mass m moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a road is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the road is to be v = 12.8 m/s (28.6 mi/h) and the radius of the curve is r = 37.0 m. At what angle should the curve be banked?
Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:
[tex]a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2[/tex]
Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So [tex] gsin\alpha = 4.43cos\alpha[/tex]
[tex]\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451[/tex]
[tex]tan\alpha = 0.451[/tex]
[tex]\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0[/tex]
Using Newton's second law and free-body diagram the angle with which the curve is to be banked is obtained as [tex]24.31^\circ[/tex].
Newton's Second Law of MotionThis problem can be analysed using a free-body diagram.
The acceleration in the horizontal direction (radial diretion) is the centripetal acceleration.
Applying Newton's second law of motion in the x-direction, we get;
[tex]\sum F = ma[/tex]
[tex]\implies N\,sin\, \theta =\frac{mv^2}{r}[/tex]
Now, applying Newton's second law of motion in the y-direction, we get;
[tex]\implies N\,cos \, \theta=mg[/tex]
Dividing both the equations, we get;
[tex]\frac{N\,sin\, \theta}{N\,cos\, \theta} =\frac{mv^2}{r\, mg}[/tex]
[tex]\implies tan\theta=\frac{v^2}{rg}=\frac{12.8^2}{37\times9.8} =0.4518[/tex]
[tex]\implies \theta=tan^{-1}(0.4518)=24.31^\circ[/tex]
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A ball is thrown upward at a speed v0 at an angle of 58.0˚ above the horizontal. It reaches a maximum height of 8.0 m. How high would this ball go if it were thrown straight upward at speed v0?
Answer:
11.245 m
Explanation:
The vertical component of the initial velocity v0 is
[tex]v_v = v_0sin58^0 = 0.848v_0[/tex]
This makes the ball reach a maximum height of 8m. If we apply the conservation law of mechanical energy, its kinetic energy is converted to potential energy when it travels to the maximum height
[tex]E_p = E_k[/tex]
[tex]mgh = mv_v^2/2[/tex]
where m is the mass and h = 8 m is the maximum vertical distance traveled, g = 9.81m/s2 is the gravitational acceleration
we can divide both sides by m
[tex]gh = v_v^2/2[/tex]
[tex](0.848v_0)^2 = 2gh = 2*9.81*8 = 156.96[/tex]
[tex]0.848v_0 = \sqrt{156.96} = 12.53[/tex]
[tex]v_0 = 12.53 / 0.848 = 14.77 m/s[/tex]
So if the ball is directed fully upward at v0 speed then we can apply the same equation to find the new H
[tex]E_p = E_k[/tex]
[tex]mgH = mv_0^2/2[/tex]
[tex]H = \frac{v_0^2}{2g} = \frac{14.77^2}{2*9.81} = \frac{218.3}{19.62} = 11.245m[/tex]
the ball would go approximately 0.454 times as high if it were thrown straight upward at speed v0 compared to when it was thrown at an angle of 58.0 degrees above the horizontal.
When a projectile is launched at an angle, its vertical motion is influenced by gravity, while its horizontal motion is independent of gravity. In this case, the ball is initially launched at an angle of 58.0 degrees above the horizontal and reaches a maximum height of 8.0 m.
If the ball were thrown straight upward at the same initial speed (v0), its vertical motion would still be influenced by gravity. In both cases, the initial vertical speed (in the upward direction) is v0. The time it takes to reach the maximum height in both scenarios would also be the same because only the vertical component of velocity affects the vertical motion.
To find how high the ball would go if thrown straight upward, you can use the following kinematic equation:
h = (v0² * sin²(θ)) / (2 * g)
Where:
h is the maximum height (which we want to find).
v0 is the initial speed (given as v0).
θ is the launch angle (58.0 degrees).
g is the acceleration due to gravity (approximately 9.81 m/s²).
First, convert the angle from degrees to radians:
θ (in radians) = 58.0 degrees * (π radians / 180 degrees) ≈ 1.01 radians
Now, plug in the values and solve for h:
h = (v0² * sin²(1.01)) / (2 * 9.81 m/s²)
h ≈ (v0² * 0.454) / 19.62 m/s²
Now, if we consider that the initial speed v0 remains the same and throw the ball straight upward (θ = 90 degrees), the equation becomes:
h_straight_up = (v0² * sin²(π/2)) / (2 * 9.81 m/s²)
h_straight_up ≈ (v0² * 1) / 19.62 m/s²
Now, compare the two expressions for h and h_straight_up:
h / h_straight_up ≈ ((v0² * 0.454) / 19.62 m/s²) / ((v0² * 1) / 19.62 m/s²)
The "v0²" terms cancel out, and you get:
h / h_straight_up ≈ 0.454 / 1
h / h_straight_up ≈ 0.454
So, the ball would go approximately 0.454 times as high if it were thrown straight upward at speed v0 compared to when it was thrown at an angle of 58.0 degrees above the horizontal.
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What is the change in temperature of a 2.50 L system when its volume is reduced to 1.00 L if the initial temperature was 298 K?
Answer:
-178.8 K
Explanation:
From Charles law,
V₁/T₁ = V₂/T₂.................... Equation 1
Where V₁ = Initial volume, T₁ = Initial Temperature, V₂ = Final volume, T₂ = Final Temperature.
Making T₂ the subject of the equation
T₂ = V₂T₁/V₁............... Equation 2
Given: V₂ = 1.00 L, V₁ = 2.5 L, T₁ = 298 K.
Substitute into equation 2
T₂ = 1.00(298)/2.5
T₂ = 119.2 K.
But,
Change in temperature = T₂ - T₁ = 119.2-298
Change in temperature = -178.8 K.
Hence the change in temperature = -178.8 K
The change in temperature of a system when its volume is reduced is calculated by utilizing Charles's law which states the volume of a gas is directly proportional to its absolute temperature under constant pressure.
Explanation:
To answer your query about the change in temperature when a system's volume is reduced, we would use Charles's law. Initially, the volume (V₁) is 2.50 L and the temperature (T₁) is 298 K. When the volume is reduced to 1.00 L (V₂), we're required to find the new temperature (T₂). According to Charles's law, V₁/T₁=V₂/T₂.
By manipulating this formula, we calculate the final temperature (T₂) as T₂=(V₂ * T₁)/V₁. Substituting the given values into this equation, we find T₂=(1.00 L * 298 K)/2.50 L. After performing the calculation, you will have the final temperature in Kelvin (K) when the volume is reduced to 1.00 L.
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If we were to construct an accurate scale model of the solar system on a football field with the Sun at one end and Neptune at the other, the planet closest to the center of the field would be (a) Earth; (b) Jupiter; (c) Saturn; (d) Uranus.
Answer:
a) Earth
Explanation:
When constructing an accurate scale model of the solar system on a football field with the Sun at one end and Neptune at the other, the planet closest to the center of the field would be Earth.
From the given options the earth is the closest planet to the center of the field. The center of the field can be considered as the sun. In reality we have mercury as the closest planet in our solar system.
A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of its acceleration.
Answer:
The acceleration of the laundry cart is 13.3 m/s²
Explanation:
Hi there!
According to Newton's second law, the sum of all forces acting on an object in a given direction is equal to the mass of the object times its acceleration in that direction:
∑F = m · a
Where:
∑F = net force acting on the cart.
m = mass of the cart.
a = acceleration.
Then, solving this equation for the acceleration:
∑F / m = a
60.0 N / 4.50 kg = a
a = 13.3 m/s²
The acceleration of the laundry cart is 13.3 m/s²
Using the formula of acceleration a = F/m, where F is the net external force and m is the mass, substituting the given values yields an acceleration of 13.33 m/s².
Explanation:According to Newton's second law, the acceleration of an object as produced by a net force is directly proportional to the net force, in the same direction as the net force, and inversely proportional to the mass of the object. This can be expressed with the formula a = F/m, where F is the net external force and m is the object's mass.
Given the mass m of the laundry cart is 4.5 kg and the net external force F on the cart is 60.0 N, we can substitute these values into the formula.
Therefore, the acceleration a = F/m = 60.0 N / 4.50 kg = 13.33 m/s².
So, the magnitude of the cart's acceleration is 13.33 m/s².
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Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so that rocket slows down at the rate of 2.05 m/s2 . A 6.50-kg instrument is hanging by a vertical wire inside a space ship.Find the force that the wire exerts on the instrument.
Answer:
R= 78.32 N
Explanation:
Given that
Acceleration ,a= 2.05 m/s²
Mass , m = 6.5 kg
The force due to acceleration
F= mass x Acceleration
F= ma
F= 6.5 x 2.05 N
F= 13.32 N
The force due to weight
F' = m g
F' = 6.5 x 10 N ( take g= 10 m/s²)
F'= 65 N
Therefore the net total force will be summation of force due to weight and force due to acceleration
R= F + F'
R= 65 + 13.32 N
R= 78.32 N
The force that the wire exerts on the instrument is R= 78.32 N
Calculation of force:
Given that
Acceleration ,a= 2.05 m/s²Mass , m = 6.5 kgThe force due to accelerationSince The force due to weight
[tex]F= ma\\\\F= 6.5 \times 2.05 N\\\\F= 13.32 N[/tex]
Now
[tex]F' = m g\\\\F' = 6.5 \times 10 N ( take\ g= 10 m/s^2)[/tex]
F'= 65 N
Now
R= F + F'
R= 65 + 13.32 N
R= 78.32 N
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Two brothers push on a box from opposite directions. If one brother pushes with a force of 86 N and the other brother pushes with a force of 67 N, what is the magnitude of the net force on the box?
Answer:
R=19 N
Explanation:
Given that forces are taking opposite to each other.
Take
F₁ = 86 N ( Towards right ,take positive)
F₂ = - 67 N ( Towards left)
Given that the angle between above two forces is 180° so the resultant can be given as
Lets take the resultant = R N
R= F₁ + F₂
Now by putting the values
R= 86 N - 67 N
R=19 N
Therefore the resultant force on the block box will be 19 N.
Final answer:
The magnitude of the net force on the box is 19 N, calculated by subtracting the smaller force from the larger force since they are directed oppositely.
Explanation:
When two brothers push on a box from opposite directions, one with a force of 86 N and the other with a force of 67 N, the net force on the box is determined by subtracting the smaller force from the larger force because they are applied in opposite directions. Therefore, the magnitude of the net force on the box can be calculated as follows:
Net Force = Larger Force - Smaller Force
Net Force = 86 N - 67 N
Net Force = 19 N
Hence, the magnitude of the net force acting on the box is 19 N.
You are 2m from one audio speaker and 2.1m from another audio speaker. Both generate the identical sine wave with a frequency of 680 Hz. At your location, what is the phase difference between the waves? Give the answer in radians, using 340m/s as the velocity of sound.
Answer:
the phase difference is 1.26 radian
Solution:
As per the question:
Distance, d = 2 m
Distance from the other speaker, d' = 2.1 m
Frequency, f = 680 Hz
Speed of sound, v = 340 m/s
Now,
To calculate the phase difference, [tex]\Delta \phi[/tex]:
Path difference, [tex]\Delta d = d' - d = 2.1 - 2 = 0.1\ m[/tex]
For the wavelength:
[tex]f\lambda = v[/tex]
where
c = speed of light in vacuum
[tex]\lambda [/tex] = wavelength
Now,
[tex]680\times \lambda = 340[/tex]
[tex]\lambda = 0.5\ m[/tex]
Now,
Phase difference, [tex]\Delta phi = 2\pi \frac{\Delta d}{\lambda}[/tex]
[tex]\Delta phi = 2\pi \frac{0.1}{0.5} = 1.26\ rad[/tex]
A circular tube is subjected to torque T at its ends. The resulting maximum shear strain in the tube is 0.005. Calculate the minimum shear strain in the tube and the shear strain at the median line of the tube section.
Answer:
The minimum shear strain in the tube is [tex]4.162\times10^{-3}[/tex]
The shear strain at the median line of the tube section is [tex]9.15\times10^{-3}[/tex]
Explanation:
Given that,
Maximum shear strain = 0.005
We need to calculate the minimum shear strain
Using formula of maximum shear strain
[tex]\gamma_{max}=\dfrac{d_{2}}{2}\times\theta[/tex]
[tex]\theta=\dfrac{2\times\gamma_{max}}{d_{2}}[/tex]
Where, [tex]\gamma_{max}[/tex]=maximum shear strain
[tex]\theta[/tex]=angle of twist
[tex]d_{2}[/tex]= diameter
Put the value into the formula
[tex]\theta=\dfrac{2\times0.005}{3}[/tex]
[tex]\theta=0.00333\ rad[/tex]
[tex]\theta=3.33\times10^{-3}\ rad[/tex]
Now, Using formula of minimum shear strain
[tex]\gamma_{min}=\dfrac{d_{1}}{2}\times\theta[/tex]
Put the value into the formula
[tex]\gamma_{min}=\dfrac{2.5}{2}\times3.33\times10^{-3}[/tex]
[tex]\gamma_{min}=0.0041625[/tex]
[tex]\gamma_{min}=4.162\times10^{-3}[/tex]
We need to calculate the shear strain at the median line of the tube section
Using formula of shear strain at the median line
[tex]\gamma=\dfrac{d_{1}+d_{2}}{2}\times\theta[/tex]
Put the value into the formula
[tex]\gamma=\dfrac{2.5+3}{2}\times3.33\times10^{-3}[/tex]
[tex]\gamma=0.0091575[/tex]
[tex]\gamma=9.15\times10^{-3}[/tex]
Hence, The minimum shear strain in the tube is [tex]4.162\times10^{-3}[/tex]
The shear strain at the median line of the tube section is [tex]9.15\times10^{-3}[/tex]
Explanation of how to calculate the minimum shear strain and the shear strain at the median line in a circular tube subjected to torque.
Shear Strain Calculations:
Minimum shear strain can be calculated using the formula: minimum shear strain = maximum shear strain / 2Shear strain at the median line of the tube section is calculated when shear strain is maximum/2, therefore it is 0.005 / 2 = 0.0025
What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 10 − 21 J ?
Answer:
404K
Explanation:
Data given, Kinetic Energy.K.E=8.37*10^-21J
Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.
This relationship can be expressed as
[tex]K.E=\frac{3}{2}KT[/tex]
where K is a constant of value 1.38*10^-23
Hence if we substitute the values, we arrive at
[tex]T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K[/tex]
converting to degree we have [tex]131^{0}C[/tex]
List the following items in order of decreasing speed, from greatest to least: (A) A wind-up toy car that moves 0.10 m in 2.5 s . (B) A soccer ball that rolls 2.5 m in 0.55 s . (C) A bicycle that travels 0.60 m in 7.5×10−2 s .(D) A cat that runs 8.0 m in 2.5 s . Enter your answer as four letters separated with commas.
Answer:
bicycle>soccer ball>cat>toy car
Explanation:
s = Distance
t = Time
Speed is given by
[tex]v=\dfrac{s}{t}[/tex]
For toy car
[tex]v=\dfrac{0.1}{2.5}=0.04\ m/s[/tex]
For soccer ball
[tex]v=\dfrac{2.5}{0.55}=4.54\ m/s[/tex]
For bicycle
[tex]v=\dfrac{0.6}{7.5\times 10^{-2}}=8\ m/s[/tex]
For cat
[tex]v=\dfrac{8}{2.5}=3.2\ m/s[/tex]
8>4.54>3.2>0.04
bicycle>soccer ball>cat>toy car
A firm wants to determine the amount of frictional torque in their current line of grindstones, so they can redesign them to be more energy efficient. To do this, they ask you to test the best-selling model, which is basically a diskshaped grindstone of mass 1.3 kg and radius 8.50 cm which operates at 700 rev/min. When the power is shut off, you time the grindstone and find it takes 41.3 s for it to stop rotating.
Answer:
0.00833542627085 Nm
Explanation:
[tex]\omega_f[/tex] = Final angular velocity = 700 rpm
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
m = Mass of stone = 1.3 kg
r = Radius = 8.5 cm
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-700\dfrac{2\pi}{60}}{41.3}\\\Rightarrow \alpha=-1.77491110372\ rad/s^2[/tex]
Torque is given by
[tex]\tau=-I\alpha\\\Rightarrow \tau=-\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=-\dfrac{1}{2}1.3\times 0.085^2\times -1.77491110372\\\Rightarrow \tau=0.00833542627085\ Nm[/tex]
The frictional torque is 0.00833542627085 Nm
Is it possible for an object to be (a) slowing down while its acceleration is increasing in magnitude; (b) speeding up while its acceleration is decreasing? In both cases, explain your reasoning.
a) Yes, if acceleration and velocity have opposite directions
b) Yes, if acceleration and velocity have same direction
Explanation:
a)
In order to answer this question, we have to keep in mind that both velocity and acceleration are vector quantities, so they have also a direction.
Acceleration is defined as the rate of change in velocity:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, the object is slowing down: this means that the magnitude of its velocity is decreasing, so
[tex]|v|<|u|[/tex]
This means that the direction of the acceleration is opposite to the direction of the velocity. Then, the magnitude of the acceleration can be increasing (this will not affect the fact that the object will slow down, but it will affect only the rate at which the object is slowing down).
b)
In this case, the object is speeding up. This means that the magnitude of its velocity is increasing, so we have
[tex]|v|>|u|[/tex]
In order for this to happen, it must be that the direction of the acceleration is the same as the direction of the velocity: therefore, this way, the magnitude of the velocity will be increasing (either in the positive or in the negative direction).
Then, the magnitude of the acceleration can be decreasing (this will not affect the fact that the object will speed up, but it will only affect the rate at which the object is speeding up).
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Two tuning forks sounding together result in a beat frequency of 3 Hz. If the frequency of one of the forks is 256 Hz, what is the frequency of the other?
Answer:
The other frequency is either 253Hz (the altered tuning frequency vibrates lower) or 259Hz(if the altered tuning frequency vibrates higher):
Explanation:
We have two frequencies: the tuning fork frequency and the altered tuning fork frequency
F1 -----tuning fork frequency
F2-----Altered tuning fork frequency
Fbeat=3
F1=256Hz
To find the alterd tuning fork frequency:
The beat frequency of any two waves is:
Fbeat =|f1 – f2|
i.e we have two choices
for frequency tied to length increase (wavelength increases as length increases, therefore frequency decrease (the altered tuning frequency vibrates lower): Fbeat = f2-f1
F2=F1-Fbeat
F2= 256-3
=253Hz
For frequency tied to length decrease wavelength increases as length increases, therefore frequency also increases(the altered tuning frequency vibrates higher) : Fbeat = -(f2-f1)
F2= Fbeat+F1
= 256+3
= 259Hz
Note that a wavelength increase means a decrease of frequency because v = fλ
If we increase the temperature in a reactor by 54degrees Fahrenheit [°F], how many degrees Celsius [°C] will the temperature increase?
Answer:
If we increase the temperature in a reactor by 54 degrees Fahrenheit [54°F], the temperature will increase by 12.22 degrees Celsius [12.22 ⁰C]
Explanation:
To determine the number of degrees Celsius the temperature will be increased, we convert from Fahrenheit to Celsius.
Converting from Fahrenheit to degree Celsius
54°F -----> °C
54 = 1.8°C + 32
54-32 = 1.8°C
22 = 1.8°C
°C = 22/1.8
= 12.22 °C
Thus, 54°F -----> 12.22 °C
Therefore, If we increase the temperature in a reactor by 54degrees Fahrenheit [54°F], the temperature will increase by 12.22 degrees Celsius [12.22 ⁰C]
12.22°C
Explanation:The temperature increase in the reactor is 54°F
Now let's convert this to °C using the following relation;
(x − 32) × 5/9 = y -------------------(i)
Where;
x is the value of the degree Fahrenheit = 54
y is the value of its corresponding degree Celsius.
Substitute x = 54 into equation (i)
(54 − 32) × 5/9 = y
(22) × 5/9 = y
Solve for y;
y = 22 x 5/9
y = 12.22°C
Therefore, the increase in temperature of the reactor in °C is 12.22
What would be the frequency of an electromagnetic wave having a wavelength equal to Earth’s diameter (12,800 km)? In what part of the electromagnetic spectrum would such a wave lie?
Answer:
f = 23.43 Hz for radio waves.
Explanation:
In this case, we need to find the frequency of an electromagnetic wave having a wavelength equal to Earth’s diameter i.e. 12,800 km. The relation between the frequency, wavelength and the speed is given by :
[tex]v=f\lambda[/tex]
Electromagnetic wave moves with a speed of light
[tex]c=f\lambda[/tex]
[tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8}{12800\times 10^3}[/tex]
f = 23.43 Hz
So, the frequency of an electromagnetic wave is 23.43 Hz. It belongs to radio waves.
Final answer:
The frequency of an electromagnetic wave with a wavelength of Earth's diameter (12,800 km) would be approximately 23.4 Hz and would lie in the extremely low frequency (ELF) portion of the electromagnetic spectrum, typically used for submarine communication.
Explanation:
The frequency of an electromagnetic wave can be calculated using the formula c = λf, where c is the speed of light (approximately 3×108 m/s), λ is the wavelength, and f is the frequency. Given the Earth’s diameter as the wavelength (12,800 km, which is 12,800,000 meters), we can rearrange the formula to solve for frequency: f = c / λ.
Plugging in the values, we get:
f = 3×108 m/s / 12,800,000 m
This results in a frequency of approximately 23.4 Hz. Electromagnetic waves with this frequency would fall in the extremely low frequency (ELF) range of the electromagnetic spectrum, which is commonly used for submarine communication due to its ability to penetrate deep into the ocean.
The following conversions occur frequently in physics and are very useful. (a) Use 1 mi = 5280 ft and 1 h = 3600 s to convert 60 mph to units of ft/s. (b) The acceleration of a freely falling object is 32 ft/s2. Use 1 ft = 30.48 cm to express this acceleration in units of m/s2. (c) The density of water is 1.0 g/cm3. Convert this density to units of kg/m3.
To solve this exercise we will define the units of conversion between each of the variables given to facilitate the calculation in each section.
[tex]1 mile = 5280ft[/tex]
[tex]1h = 3600s[/tex]
[tex]1 ft = 30.48cm[/tex]
[tex]1 m = 100cm[/tex]
[tex]1kg = 1000g[/tex]
PART A ) For this part we have 60mph to ft/s, then
[tex]60mph = 60\frac{miles}{hour} (\frac{5280ft}{1mile})(\frac{1h}{3600s})[/tex]
[tex]60mph = 88ft/s[/tex]
PART B) Convert from [tex]ft/s^2[/tex] to [tex]m/s^2[/tex]
[tex]32ft/s^2 = 32 \frac{ft}{s^2} (\frac{30.48cm}{1ft})(\frac{1m}{100cm})[/tex]
[tex]32ft/s^2 = 9.7536m/s^2[/tex]
PART C) Convert [tex]g/cm^3[/tex] to [tex]kg/m^3[/tex]
[tex]1 g/cm^3 = 1\frac{g}{cm^3}(\frac{1kg}{1000g})(\frac{100cm}{1m})^3[/tex]
[tex]1 g/cm^3 = 1000kg/m^3[/tex]
Based on the conversion factors, the conversions are as follows:
60 mph = 88 ft/s32 ft/s² = 9.75 m/s²1 g/cm³ = 1000 Kg/m³What are the conversions for the given units?Conversions are used when converting between different units measuring the same quantity.
a. To convert 60 mph to ft/s:
1 mi = 5280 ft 1 h = 360060 mph = 60 * 5280/3600
60 mph = 88 ft/s
b. To convert 32 ft/s² to m/s
1 ft = 30.48 cm = 0.3048 m
32 ft/s² = 32 * 0.3034
32 ft/s² = 9.75 m/s²
c. To convert g/cm³ to kg/m³
1 g = 0.001 kg
1 cm³ = 0.000001 m³
1.0 g/cm3 = 1.0 * 0.001 kg/0.000001 m³
1 g/cm³ = 1000 Kg/m³
Therefore, the conversions are as follows:
60 mph = 88 ft/s32 ft/s² = 9.75 m/s²1 g/cm³ = 1000 Kg/m³Learn more about unit conversions at: https://brainly.com/question/174910
Two 1.0 g balls are connected by a 2.0-cm-long insulating rod of negligible mass. One ball has a charge of +10 nC, the other a 4 charge of - 10 nC. The rod is held in a 1.0 * 10 N/C uniform electric field at an angle of 30° with respect to the field, then released. What is its initial angular acceleration?
Answer:
α = 5 10⁻³ rad / s²
Explanation:
For this exercise we can use Newton's second law for rotational movement, where the force is electric
τ = I α
Where the torque is
τ = F x r = F r sin θ
Strength is
F = q E
The moment of inertia of a small ball, which we approximate to a point is
I = m r²
We replace
2 (q E) r sin θ = 2m r² α
The number 2 is because the two forces create the same torque
α = q E sin θ / m r
Let's reduce the magnitudes to the SI system
m = 1.0g = 1.0 10⁻³ kg
L = 2.0 cm = 2.0 10⁻² m
q = 10 nc = 10 10⁻⁹ C
E = 1.0 10 N / C
r = L / 2
r = 1.0 10⁻² m
Let's calculate
α = 10 10⁻⁹ 1.0 10 sin 30 / 1.0 10⁻³ 1.0 10⁻²
α = 5 10⁻³ rad / s²
The initial angular acceleration of the system is 0.25 radians per second squared.
To find the initial angular acceleration of the system when the charged balls are released in the uniform electric field, we can use the concept of torque. The torque experienced by the system will cause it to rotate, resulting in angular acceleration.
The torque (τ) experienced by an electric dipole in an electric field is given by the formula:
τ = p * E * sin(θ)
Where:
- τ is the torque.
- p is the electric dipole moment, which is the product of the charge (q) and the separation (d) between the charges on the dipole: p = q * d.
- E is the electric field strength.
- θ is the angle between the electric field direction and the dipole moment direction.
In this case, we have:
- q1 = +10 nC (positive charge)
- q2 = -10 nC (negative charge)
- d = 2.0 cm = 0.02 m
- E = 1.0 * 10^3 N/C (1.0 * 10^3 N/C = 1.0 * 10^3 V/m, as 1 N/C = 1 V/m)
- θ = 30°
First, calculate the electric dipole moment (p):
p = q * d = (+10 * 10^-9 C) * (0.02 m) = 2 * 10^-10 C·m
Now, calculate the torque (τ):
τ = p * E * sin(θ) = (2 * 10^-10 C·m) * (1.0 * 10^3 V/m) * sin(30°)
To calculate sin(30°), we can use its exact value: sin(30°) = 1/2.
τ = (2 * 10^-10 C·m) * (1.0 * 10^3 V/m) * (1/2) = 1 * 10^-7 N·m
Now, we can find the moment of inertia (I) of the system. Since we have two balls rotating at a fixed distance (d), we can use the formula for a simple pendulum's moment of inertia:
I = 2 * m * d^2
Where:
- m is the mass of each ball (1.0 g = 0.001 kg)
- d is the separation (0.02 m)
I = 2 * (0.001 kg) * (0.02 m)^2 = 4 * 10^-7 kg·m^2
Now, we can calculate the initial angular acceleration (α) using the equation:
τ = I * α
α = τ / I = (1 * 10^-7 N·m) / (4 * 10^-7 kg·m^2) = 0.25 rad/s^2
So, the initial angular acceleration of the system is 0.25 radians per second squared.
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A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min at 80.0 km/h, 12.0 min at 100 km/h, and 45.0 min at 40.0 km/h and spends 15.0 min eating lunch and buying ga .
(a) Determine the average peed for t he trip.
(b) Determine the d istance between the initial and final cit ies along the route.
Answer:
a.52.9 km/h
b.90 km
Explanation:
We are given that
[tex]v_1=89km/h[/tex]
[tex]t_1=30min[/tex]
[tex]v_2=100km/h[/tex]
[tex]t_2=12min[/tex]
[tex]v_3=40km/h[/tex]
[tex]t_3=45 min[/tex]
Time spend on eating lunch and buying ga=15 min.
a.Total time=30+12+45+15=102 minute=[tex]\frac{102}{60}=1.7 hour[/tex]
1 hour=60 minutes
Distance=[tex]speed\times time[/tex]
[tex]d_1=v_1\times t_1=80\times\frac{30}{60}=40km[/tex]
[tex]d_2=100\times \frac{12}{60}=20 km[/tex]
[tex]d_3=40\times \frac{45}{60}=30 km[/tex]
Total distance=[tex]d_1+d_2+d_3=40+20+30=90km[/tex]
Average speed=[tex]\frac{total\;speed}{total\;time}[/tex]
Using the formula
Average speed=[tex]\frac{90}{1.7}=52.9Km/h[/tex]
b.Total distance between the initial and final city lies along the route=90 km
In the Bohr model the hydrogen atom consists of an electron in a circular orbit of radius a 0 = 5.29 × 10 − 11 m around the nucleus. Using this model, and ignoring relativistic effects, what is the speed of the electron?
To solve this problem we will apply the concept of balance of Forces in the body. For such an effect the centripetal force must be equivalent to the electrostatic force of the body, therefore
[tex]F_c = F_e[/tex]
[tex]\frac{mv^2}{r} = k \frac{q_pq_e}{r^2}[/tex]
Here
m = Mass of electron
r = Distance between them
k = Coulomb's constant
[tex]q_p[/tex] = Charge of proton
[tex]q_e[/tex] = Charge of electron
v = Velocity
Rearranging to find the velocity we have that,
[tex]v^2 = \frac{kq_pq_e}{mr}[/tex]
[tex]v = \sqrt{\frac{kq_pq_e}{mr}}[/tex]
Replacing,
[tex]v = \sqrt{\frac{(9*10^9)(1.6*10^{-19})(1.6*10^{-19})}{(9.1*10^{-31})(5.29*10^{-11})}}[/tex]
[tex]v = 2.19*10^6m/s[/tex]
Therefore the speed of the electron is [tex]2.19*10^6m/s[/tex]
Now the same particle is removed from the thread and placed over the center of a charged plate. Are there any conditions under which it is possible for the particle to be suspended in the air above the plate? Show any relevant calculations and explain your reasoning.
Answer:
changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.
Explanation:
When the particle is removed from the wire, friction can be electrically charged, either with negative charges (extra electrons) or with positive charge by electron removal, in this case when the particle is between the condenser plates it experiences a force due to the electric field given by
ΔV = E d
Where ΔV is the potential difference, d the distance between the plates and E the electric field.
In these cases we can use Newton's second law, where the acceleration is zero
[tex]F_{e}[/tex] –W + B = 0
F_{e} = W –B
q E = mg - ρ_air g V
dodne B is the hydrostatic thrust
if we know the density of the particular
ρ_particle = m / V
m = ρ_particle V
We replace
q E = g v (ρ_particle - ρ_air)
Therefore, by changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.