Answer:
[tex]E=0[/tex] at r < R;
[tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex] at 2R > r > R;
[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex] at r >= 2R
Explanation:
Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,
[tex]\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}[/tex] (integral over a closed surface)
where,
[tex]E[/tex] = Electric field
[tex]Q_{enclosed}[/tex] = charged enclosed within the closed surface
[tex]\epsilon[/tex] = permittivity of free space
Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .
at r < R,
[tex]Q_{enclosed}[/tex] = 0 and hence [tex]E[/tex] = 0 (since the sphere is conducting, all the charges get repelled towards the surface)
at 2R > r > R,
[tex]Q_{enclosed}[/tex] = Q,
therefore,
[tex]E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}[/tex]
(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to [tex]4\pi r^{2}[/tex])
or, [tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex]
at r >= 2R
[tex]Q_{enclosed}[/tex] = 2Q
Hence, by similar calculations, we get,
[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex]
The electric field inside the solid conducting sphere is zero. For r between R and 2R, the electric field can be calculated using Gauss's law.
Explanation:The electric field inside a solid conducting sphere is zero. Thus, the magnitude of the electric field in the region 0<r<R is zero.
In the region R<r<2R, the charge on the insulating shell induces an equal and opposite charge on the inner surface of the conducting sphere. Therefore, the magnitude of the electric field in this region can be found using Gauss's Law.
The electric field magnitude is given by:
E = Q / (4πε0 r2)
where Q is the charge on the insulating shell, ε0 is the vacuum permittivity, and r is the distance from the center of the sphere.
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If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?
Answer:
[tex]heigth=2.86m[/tex]
Explanation:
Given data
time=0.19 s
distance=1.6 m
To find
height
Solution
First we need to find average velocity
[tex]V_{avg}=\frac{distance}{time}\\V_{avg}=\frac{1.6m}{0.19s}\\V_{avg}=8.42m/s[/tex]
Also we know that average velocity
[tex]V_{avg}=(V_{i}+V_{f})/2\\[/tex]
Where
Vi is top of window speed
Vf is bottom of window speed
Also we now that
[tex]V_{f}=V_{i}+gt\\V_{f}=V_{i}+(9.8)(0.19)\\V_{f}=V_{i}+1.862[/tex]
Substitute value of Vf in average velocity
So
[tex]V_{avg}=(V_{i}+V_{f})/2\\where\\V_{f}=V_{i}+1.862\\and\\V_{avg}=8.42m/s\\So\\8.42m/s=(V_{i}+V_{i}+1.862)/2\\2V_{i}+1.862=16.84\\V_{i}=(16.84-1.862)/2\\V_{i}=7.489m/s\\[/tex]
Vi is speed of balloon at top of the window
Now we need to find time
So
[tex]V_{i}=gt\\t=V_{i}/g\\t=7.489/9.8\\t=0.764s[/tex]
So the distance can be found as
[tex]distance=(1/2)gt^{2}\\ distance=(1/2)(9.8)(0.764)^{2}\\ distance=2.86m[/tex]
To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight. Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can solve for the initial height to be approximately 0.01485 m or 14.85 cm above the top of the window.
Explanation:To determine the height from which the balloon was dropped, we can use the equation for vertical motion: h = 0.5*g*t^2, where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Given that the balloon takes 0.19 s to cross the 1.6 m high window, we can plug in these values into the equation to solve for the initial height:
1.6 = 0.5*9.8*(0.19)^2
Simplifying the equation, we find that the balloon was dropped from a height of approximately 0.01485 m or 14.85 cm above the top of the window.
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At the instant that you fire a bullet horizontally from a rifle, you drop a bullet from the height of the gun barrel. If there is no air resistance, which bullet hits the level ground first? Explain.
Answer:
Both hit at the same time
Explanation:
If air resistant is ignored, then gravitational acceleration g is the only thing that affect the bullets vertical motion, no matter what their horizontal motion is. Since both bullets are starting from rest, vertically speaking their speeds are 0 initially, they are both subjected to the same acceleration g, then they travel at the same rate and would reach the ground at the same time.
In building a particle accelerator, you manage to produce a uniform electric field of magnitude 6.03 × 10 5 N/C in one 35.5 cm section. Calculate the magnitude of the electric potential difference across the length of the accelerator's section. How much work is required to move a proton through the section?
Answer:
V = 2.14×10⁵ V.
W = 3.424×10⁻¹⁴ J.
Explanation:
Electric Potential: This can be defined as the work done in bringing a unit positive charge from infinity to that point, against the action of a field.
The S.I unit is V.
The expression containing electric potential, distance and electric field is given as,
V = E×r .............. Equation 1
Where V = Electric potential difference across the length of the accelerator's section, E = Electric Field, r = Length of the section.
Given: E = 6.03×10⁵ N/C, r = 35.5 cm = 0.355 m.
Substitute into equation 1
V = 6.03×10⁵×0.355
V = 2.14065×10⁵ V.
V ≈ 2.14×10⁵ V.
amount of Work required to move a proton through the section is given as,
W = qV ............... Equation 2
Where W = work required to move a proton through the section, q = charge on a proton V = Electric potential.
Given: V = 2.14×10⁵ V, q = 1.60 x 10⁻¹⁹ C.
Substitute into equation 2
W = (2.14×10⁵)(1.60 x 10⁻¹⁹)
W = 3.424×10⁻¹⁴ J.
What is the maximum number of 7.00 μF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC ?
Final answer:
To maintain the total charge below 953 µC for capacitors connected in parallel to a 3.00 V battery, a maximum of 45 capacitors, each with a capacitance of 7.00 µF, can be used. This is calculated by dividing the total charge by the charge per capacitor (953 µC / 21 µC per capacitor).
Explanation:
To find out the maximum number of 7.00 µF capacitors that can be connected in parallel while keeping the total charge under 953 µC, we use the formula for charge on a capacitor, Q=CV, where C is the capacitance and V is the voltage. Since the capacitors would be connected in parallel, the voltage across each capacitor would be the same as the voltage of the battery, which is 3.00 V.
The total charge Q for one capacitor is given by:
Q = C × V = 7.00 µF × 3.00 V = 21.00 µC per capacitor.
To find out how many such capacitors we could connect in parallel without exceeding the total charge of 953 µC, we divide the total permissible charge by the charge per capacitor:
Number of capacitors = Total charge / Charge per capacitor = 953 µC / 21.00 µC per capacitor ≈ 45.38.
Since we can't have a fraction of a capacitor, the maximum number you can use is 45 capacitors.
The maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored under 953 μC is 45 capacitors.
To determine the maximum number of 7.00 µF capacitors that can be connected in parallel with a 3.00 V battery while keeping the total charge stored within the capacitor array under 953 μC, follow these steps:
First, recall the formula for the charge stored in a capacitor: Q = C × V, where Q is the charge, C is the capacitance, and V is the voltage.Each 7.00 µF capacitor connected to a 3.00 V battery will store a charge of: Q = 7.00 µF × 3.00 V = 21.00 μCTo find the maximum number of capacitors (n) that can be connected while keeping the charge under 953 μC, set up the inequality. n × 21.00 μC < 953 μCSolving for n, we get: n < 953 μC / 21.00 μC ≈ 45.38Since the number of capacitors must be an integer, the maximum number of 7.00 µF capacitors that can be connected in parallel is 45 capacitors.
Consider a bicycle that has wheels with a circumference of 1 m. What is the linear speed of the bicycle when the wheels rotate at 2 revolutions per second?
Answer:
V= 12.56 m/s
Explanation:
Given that
R= 1 m
The angular speed ,ω = 2 rev/s
We know that 1 rev = 2 π rad
Therefore
ω = 4 π rev/s
The linear speed V is given as
V= ω x R
V = 4 π x 1 m/s
V= 4 π m/s
Therefore the speed of the bicycle will be 4 π m/s
We know that , π = 3.14
V= 12.56 m/s
The linear speed will be 12.56 m/s
A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m/s four seconds later. How high was the cliff?
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
[tex]v = v_0 -gt[/tex]
Here,
v = Final velocity
[tex]v_0[/tex] = Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
[tex]-30 = v_0 -9.8(4)[/tex]
[tex]v_0 = 9.2m/s[/tex]
Now the position can be calculated as,
[tex]y = h +v_0t -\frac{1}{2}gt^2[/tex]
When it has the ground, y=0 and the time is t=4s,
[tex]0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2[/tex]
[tex]h = 41.6m[/tex]
Therefore the cliff was initially to 41.6m from the ground
The height of the cliff from which the ball was thrown is approximately 78.4 meters, as determined by the laws of Physics and specifically equations describing motion in a gravitational field.
Explanation:To calculate the height of the cliff you can use the laws of Physics, specifically the equations that deal with motion in a gravitational field. The question states that the ball is thrown upward and it hits the ground with a speed of 30 m/s. Since the ball is thrown upwards, the initial speed is 0 m/s. The only force acting on the ball is gravity, hence, the acceleration is 9.8 m/s² (downward). So let's put these values into motion equation. Our known values are final speed (v) = 30 m/s, acceleration (a) = -9.8 m/s² (we take negative as it’s downward motion) and time (t) = 4 s. We want to find out the distance (s). We can use the second equation of motion: s = ut + 0.5*at². Substituting the known values, we get 0*4 + 0.5*-9.8*4² = -78.4 m. The negative sign indicates that the ball is below the starting point, which makes sense since it fell off a cliff. So, the height of the cliff is 78.4 meters.
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A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is fixed in place at the bottom of the incline. If the inclined plane makes an angle θ of 35.0 ∘ with the horizontal, what is the magnitude of the acceleration of the box when it is 61.0 cm from the bottom of the incline?
The magnitude of the acceleration of the box is [tex]a=6.41\ m/s^2.[/tex]
First, we need to calculate the force between the two charges using Coulomb's law:
[tex]\[ F = \frac{qQ}{4\pi\epsilon_0r^2} \][/tex]
Given that [tex]\( q = +2.50 \, \mu C = 2.50 \times 10^{-6} \, C \), \( Q = +75.0 \, \mu C = 75.0 \times 10^{-6} \, C \)[/tex], and [tex]\( r = 61.0 \, cm = 0.610 \, m \)[/tex], we can plug these values into the equation:
[tex]\[ F = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.610 \, m)^2} \][/tex]
Solving this, we get the force [tex]\( F \)[/tex] in newtons.
Next, we need to find the component of this force parallel to the incline, which is given by:
[tex]\[ F_{\parallel} = F \sin(\theta) \][/tex]
However, since the box is on an incline, we use [tex]\( \tan(\theta) \)[/tex] instead of [tex]\( \sin(\theta) \)[/tex] to find the acceleration:
[tex]\[ F_{\parallel} = F \tan(\theta) \][/tex]
Given [tex]\( \theta = 35.0^{\circ} \)[/tex], we can calculate [tex]\( F_{\parallel} \)[/tex].
Finally, the acceleration \( a \) of the box is given by Newton's second law:
[tex]\[ a = \frac{F_{\parallel}}{m} \][/tex]
where[tex]\( m = 495 \ , g = 0.495 \, kg \)[/tex].
Combining all the steps, we have:
[tex]\[ a = \frac{qQ}{4\pi\epsilon_0mr^2} \tan(\theta) \][/tex]
Plugging in the values, we get:
[tex]\[ a = \frac{(2.50 \times 10^{-6} \, C)(75.0 \times 10^{-6} \, C)}{4\pi(8.85 \times 10^{-12} \, F/m)(0.495 \, kg)(0.610 \, m)^2} \tan(35.0^{\circ}) \][/tex]
Calculating the value of [tex]\( a \)[/tex], we find the magnitude of the acceleration of the box.
[tex]a=6.41\ m/s^2[/tex]
Problem 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth.
The actual mass of the Earth (5.97 \times 10^24 kg).
To find the mass of the Earth using the given information about the moon's orbit and revolution period, we can use Kepler's third law, which states that the square of the orbital period of a planet or moon is proportional to the cube of its average distance from the planet or sun.
Let's call the mass of the Earth M and the mass of the moon m. The distance between the Earth and the moon is R = 238,910 mi.
According to Kepler's third law, we have:
(27.32 days)^2 = (238,910 mi)^3 / (M + m)
(38.79 days)^2 = (6,022,140,778 mi)^3 / M
(where 38.79 days is the sidereal period of Earth's rotation)
Dividing these two equations gives:
(38.79 days)^2 / (6,022,140,778 mi)^3 = (27.32 days)^2 / (238,910 mi)^3
Simplifying this expression:
M = (6,022,140,778 mi)^3 / [(38.79 days)^2 - (27.32 days)^2 / (1 + m/M)]^(1/3)
Since we know that m/M is very small (the mass of the moon is only about 1/81 of the mass of the Earth), we can approximate m/M as zero in this equation:
M = (6,022,140,778 mi)^3 / (38.79 days)^2 = 5.97 \times 10^24 kg
This is very close to the actual mass of the Earth (5.97 \times 10^24 kg).
A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa). How many electrons are transferred? Avogadro’s number is 6.022 × 1023 /mol, and the elemental charge is 1.602 × 10−19 C.
Answer:
Number of electrons, [tex]n=9.98\times 10^{19}[/tex]
Explanation:
A strong lightning bolt transfers an electric charge of about 16 C to Earth, q = 16 C
We need to find the number of electrons that transferred. Let there are n electrons transferred. It is given by using quantization of electric charge as :
q = ne
[tex]n=\dfrac{q}{e}[/tex]
e is elemental charge
[tex]n=\dfrac{16}{1.602\times 10^{-19}}[/tex]
[tex]n=9.98\times 10^{19}[/tex]
So, there are [tex]9.98\times 10^{19}[/tex] electrons that gets transferred. Hence, this is the required solution.
A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).
A strong lightning bolt transfers an electric charge of about 16 C to Earth (or vice versa).
What is the electric charge?Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.
We want to calculate the number of electrons that have a charge of 16 C. We have to consider the following relationships:
The charge of 1 electron is 1.602 × 10⁻¹⁹ C.There are 6.022 × 10²³ electrons in 1 mole (Avogadro's number).16 C × 1 electron/1.602 × 10⁻¹⁹ C = 1.0 × 10²⁰ electron
1.0 × 10²⁰ electron × 1 mol/6.022 × 10²³ electron = 1.7 × 10⁻⁴ mol
A strong lightning bolt that transfers an electric charge of about 16 C to Earth, transfers 1.0 × 10²⁰ electrons (1.7 × 10⁻⁴ moles of electrons).
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You are going to an outdoor concert, and you'll be standing near a speaker that emits 60 W of acoustic power as a spherical wave. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?
Answer:
[tex]r=44m[/tex]
Explanation:
β is calculated as:
[tex]\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2} )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}[/tex]
The distance r is defined as the radius of spherical wave.solve for r
We have
[tex]I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m[/tex]
Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 m/s relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?
Answer:
Total distance does the grasshopper travel before the cars hit is 150 m
Explanation:
Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is
[tex]t=\frac{x}{v}[/tex]
where x is the distance traveled with velocity v
[tex]t=\frac{100}{10}\\t=10 sec[/tex]
The insect is moving through this time period with a constant velocity of 15 m/s
The distance traveled by grasshopper is
[tex]distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m[/tex]
The grasshopper travels a total distance of 2600 m before the cars collide.
Explanation:We can solve this problem by calculating the time it takes for the cars to collide. Since the cars are moving towards each other at a combined speed of 20 m/s (10 m/s + 10 m/s), and they start 200 m apart, it will take them 200 m / 20 m/s = 10 seconds to collide.
During these 10 seconds, the grasshopper keeps jumping back and forth with a velocity of 15 m/s relative to the ground. To find the total distance the grasshopper travels, we need to calculate the number of jumps the grasshopper can make in 10 seconds. Since the grasshopper jumps the instant it lands, the number of jumps is equal to the number of times the grasshopper crosses from one car to the other. Given that the grasshopper has a velocity of 15 m/s, we divide the total distance the grasshopper travels by this velocity to find the number of jumps: 200 m / (15 m/s) = 13.33 jumps.
The grasshopper is not able to make a fraction of a jump, so we take only the whole number of jumps that the grasshopper can make: 13 jumps. Therefore, the total distance the grasshopper travels before the cars hit is 13 jumps · 200 m per jump = 2600 m.
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A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude of 86.0 m and points in a direction 42.0° above the negative x-axis. a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .
b) Using the component method of vector addition, find the magnitude of the vector C⃗ .
c) Using the component method of vector addition, find the direction of the vector C
Answer with Step-by -step explanation:
We are given that
b.[tex]\mid A\mid=46 m[/tex]
[tex]\theta=20^{\circ}[/tex] below the positive x-axis
Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=[tex]x=360-20=340^{\circ}[/tex]
x-component of vector A=[tex]A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24[/tex]
y-Component of vector A=[tex]A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64[/tex]
Magnitude of vector B=86 m
The vector B makes angle with positive x- axis=[tex]x'=42^{\circ}[/tex]
x-component of vector B=[tex]B_x=86cos42=63.64[/tex]
y-Component of vector B=[tex]B_y=86sin42=57.62[/tex]
Vector A=[tex]A_xi+A_yj=43.24i-15.64j[/tex]
Vector B=[tex]B_xi+B_yj=63.64i+57.62j[/tex]
Vector C=A+B
Substitute the values
[tex]C=43.24i-15.64j+63.64i+57.62j[/tex]
[tex]C=106.88i+41.98j[/tex]
c.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}[/tex]
The direction of the vector C=21.5 degree
To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, start by drawing the x and y axes. Use the component method of vector addition to find the magnitude of C⃗. Use the inverse tangent function to find the direction of C⃗.
Explanation:To sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗, we start by drawing the x and y axes. Vector A⃗ has a magnitude of 46.0 m and points 20.0° below the positive x-axis, so we draw A⃗ starting from the origin and make an angle of 20.0° with the positive x-axis in a downward direction. Vector B⃗ has a magnitude of 86.0 m and points 42.0° above the negative x-axis, so we draw B⃗ starting from the origin and make an angle of 42.0° with the negative x-axis in an upward direction. To find the vector C⃗=A⃗+B⃗, we add the x-components and the y-components of A⃗ and B⃗ separately. Then we use the Pythagorean theorem to find the magnitude of C⃗ and the inverse tangent function to find the direction of C⃗ in relation to the positive x-axis.
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A vector has an x component of -24.0 units and a y component of 40.4 units. Find the magnitude and direction of this vector.
Answer:
The magnitude is 47.0 and the direction is 120.7° above the x-axis counterclockwise.
Explanation:
Hi there!
The magnitude of any vector is calculated as follows:
[tex]|v| = \sqrt{x^{2} + y^{2} }[/tex]
Where:
v = vector.
x = x-component of v.
y = y-component of v.
Then, the magnitude of this vector will be:
[tex]|v| = \sqrt{24.0^{2} + 40.4^{2} } = 47.0[/tex]
The magnitude is 47.0
The direction can be found by trigonometry (see attached figure).
According to the trigonometry rules of right triangles (as the one in the figure):
cos θ = adjacent side to the angle θ / hypotenuse
In this case:
adjacent side = x-component of v.
hypotenuse = magnitude of v.
Then:
cos θ = 24.0 / 47.0
θ = 59.3°
Measured in counterclockwise:
180° - 59.3° = 120.7°
The magnitude is 47.0 and the direction is 120.7° above the x-axis counterclockwise.
A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured ������� – ��������� with respect to the positive � − ����. What are the magnitude and angle (measured ������� – ��������� with respect to the positive � − ����) of the equilibrant vector? Please show all steps in your calculations
Answer:
223 degree
Explanation:
We are given that
Magnitude of resultant vector= 8 units
Resultant vector makes an angle with positive -x in counter clockwise direction
[tex]\theta=43^{\circ}[/tex]
We have to find the magnitude and angle of the equilibrium vector.
We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.
Therefore, magnitude of equilibrium vector=8 units
x-component of a vector=[tex]v_x=vcos\theta[/tex]
Where v=Magnitude of vector
Using the formula
x-component of resultant vector=[tex]v_x=8cos43=5.85[/tex]
y-component of resultant vector=[tex]v_y=vsin\theta=8sin43=5.46[/tex]
x-component of equilibrium vector=[tex]v_x=-5.85[/tex]
y-component of equilibrium vector=[tex]-v_y=-5.46[/tex]
Because equilibrium vector lies in III quadrant
[tex]\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}[/tex]
The angle [tex]\theta'[/tex]lies in III quadrant
In III quadrant ,angle =[tex]\theta'+180^{\circ}[/tex]
Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree
The main waterline into a tall building has a pressure of 90 psia at 16 ft elevation below ground level. Howmuch extra pressure does a pump need to add to ensure a waterline pressure of 30 psia at the top floor 450 ft above ground?
Answer:
Explanation:
Given
initial Pressure [tex]P_1=90\ psia[/tex]
elevation [tex]z_1=16\ ft[/tex]
Final Pressure [tex]P_2=30\ psia[/tex]
elevation [tex]z_2=450\ ft[/tex]
Pressure after Pumping(pump inlet pressure ) is given by
[tex]P_{after\ pump}=P_{top}+\Delta P[/tex]
[tex]\Delta P=\rho gh[/tex]
where [tex]\rho [/tex]=density of water[tex](62.2\ lbm\ft^3)[/tex]
g=acceleration due to gravity[tex](32.2\ ft/s^2)[/tex]
h=elevation
[tex]\Delta P=62.2\times 32.2\times (16+450)\times \frac{1\ lbf}{32.174\ lbm\ ft}[/tex]
[tex]\Delta P=28,985\ lbf/ft^2[/tex]
[tex]\Delta P=201.3\ lbf/in.^2[/tex]
[tex]P_{after\ pump}=30+201.3=231.3\ lbf/in.^2[/tex]
Pressure required to be applied
[tex]\Delta P_{pump}=P_{after\ pump}-P_{bottom}[/tex]
[tex]\Delta P_{pump}=231.3-90=141.3\ psi[/tex]
If two such generic humans each carried 2.5 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 600 weight? *An average human weighs about 600 r= .............. km
Answer:
They would be [tex]r=9.7\,km [/tex] apart
Explanation:
Electric force between two charged objects is:
[tex] F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}[/tex] (1)
With q1, q2 the charges of the humans, r the distance between them and k the constant [tex] k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}[/tex], so if we want the electric force between them will be equal to their 600 N weight, we should make W=Fe=600 N on (1):
[tex]600=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} [/tex]
solving for r:
[tex]r^{2}=k\frac{\mid q_{1}q_{2}\mid}{600}[/tex]
[tex] r=\sqrt{k\frac{\mid q_{1}q_{2}\mid}{600}}[/tex]
[tex]r=\sqrt{(9.0\times10^{9})\frac{\mid(-2.5)(2.5)\mid}{600}} =9682 m[/tex]
[tex]r=9.7\,km [/tex]
Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.
Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From
Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the string?
a.There is no energy stored in the string: The string will remain straight for all subsequent times.
b.Energy will flow into the string, causing the standing wave to form at a later time.
c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.
d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Answer:
Explanation:
=Asin(kx−ωt). This general mathematical form can represent the displacement of a string, or the strength of an electric field, or the height of the surface of water, or a large number of other physical waves!
Part C Find ye(x) and yt(t). Remember that yt(t) must be a trig function of unit amplitude. Express your answers in terms of A, k, x, ω, and t. Separate the two functions with a comma. Use parentheses around the argument of any trig functions.
Part E At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)? Part G From
Part F we know that the string is perfectly straight at time t=π2ω. Which of the following statements does the string's being straight imply about the energy stored in the strJHJMNMMUJJHTGGHing?
a.There is no energy stored in the string: The string will remain straight for all subsequent times.
b.Energy will flow into the string, causing the standing wave to form at a later time.
c.Although the string is straight at time t=π2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string.
d.The total mechanical energy in the string oscillates but is constant if averaged over a complete cycle.
Here, the spatial and temporal parts of the wave function are A sin(kx) and sin(ωt), respectively. The displacement of the string at x=0 is 0. Although the string is straight at time t=π/2ω, there is still energy stored in it due to parts of the string having nonzero velocity.
Explanation:Part C: The spatial part of the wave function, ye(x), can be expressed as A sin(kx), and the temporal part, yt(t), is represented by sin(ωt). These expressions are valid for a wave of unit amplitude.
Part E: At position x=0, the displacement of the string is 0. This occurs because, for the wave equation, when inputting x=0 into the spatial part, the value for y becomes 0.
Part G: The correct answer is option c. Although the string is straight at time t=π/2ω, parts of the string have nonzero velocity. Therefore, there is energy stored in the string. Option d is incorrect as the total mechanical energy in a perfectly straight string does not oscillate, but remains constant.
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What are the two angles between the direction of the current and the direction of a uniform 0.0400 T magnetic field for which the magnetic force on the wire has magnitude 0.0250 N?
Answer:
The two values of θ are 41.03° and 138.97°.
Explanation:
The force on a current carrying wire is given by the following equation:
[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]
The cross-product can be written with a sine term:
[tex]F = ILB\sin(\theta)\\0.025 = IL(0.04)\sin(\theta)\\\sin(\theta) = \frac{0.025}{0.04IL}\\\theta = \arcsin(\frac{0.025}{0.04IL})[/tex]
If we assume that the wire is 0.28 m long and the current is 3.40 A, then sin(θ) becomes 0.6565.
Finally, the two values of θ are 41.03° and 138.97°.
A long string is pulled so that the tension in it increases by a factor of four. If the change in length is negligible, by what factor does the wave speed change?
To solve this problem we will apply the concepts related to wave velocity as a function of the tension and linear mass density. This is
[tex]v = \sqrt{\frac{T}{\mu}}[/tex]
Here
v = Wave speed
T = Tension
[tex]\mu[/tex] = Linear mass density
From this proportion we can realize that the speed of the wave is directly proportional to the square of the tension
[tex]v \propto \sqrt{T}[/tex]
Therefore, if there is an increase in tension of 4, the velocity will increase the square root of that proportion
[tex]v \propto \sqrt{4} = 2[/tex]
The factor that the wave speed change is 2.
A plane flies 1.4hours at 110mph on a bearing of 40degrees.It then turns and flies 1.7hours at the same speed on a bearing of 130degrees.How far is the plane from its starting point?
Answer:
Distance from starting point= 242.249871 miles
Explanation:
Distance covered on bearing of 40 degree=a= [tex]1.4*110\ mph[/tex]
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=[tex]1.7*110\ mph[/tex]
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=[tex]130-40[/tex]
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=[tex]\sqrt{a^2+b^2}[/tex]
Distance from starting point=[tex]\sqrt{154^2+187^2}[/tex]
Distance from starting point= 242.249871 miles
Answer:
Distance from starting point= 242.249871 miles
Distance covered on bearing of 40 degree=a=
Distance covered on bearing of 40 degree=a=154 miles
Distance covered on bearing of 130 degree=b=
Distance covered on bearing of 40 degree=b=187 miles
Angle between bearing=
Angle between bearing=90 degree
With the angle of 90 degree, Distance from starting point can be calculated from Pythagoras theorem.
Distance from starting point=
Distance from starting point=
Distance from starting point= 242.249871 miles
I got this from the other person so mark them brainliest
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, downward. (a) How many electrons were added to or removed from the honeybee? (b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)? (c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?
The ratio of the electric force on the bee to the bee’s weight (Fe/Fg) is; 2.041 * 10⁻⁶
The electric field strength is; 4.9 * 10⁷ N/C
What is the electric field strength?
We are given;
Mass of bee; m = 0.12g = 0.00012 kg
Charge acquired by the bee q = +24 pC = 24 * 10⁻¹² C
Electric field; E = 100 N/C
B) The force F on a charge in electric field E is given by:
F = qE
F = 24 * 10⁻¹² * 100
F = 24 * 10⁻¹⁰ N
Weight is;
F_g = mg
F_g = 0.00012 * 9.8
F_g = 11.76 × 10⁻⁴ N
Ratio of force to weight is;
F_e/F_g = (24 * 10⁻¹⁰)/(11.76 × 10⁻⁴)
F_e/F_g = 2.041 * 10⁻⁶
C) To get the electric field strength, we will use the formula;
E' = mg/q
E' = (0.00012 * 9.8)/(24 * 10⁻¹²)
E' = 4.9 * 10⁷ N/C
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Final answer:
The honeybee gained 15,000 electrons. The ratio of the electric force on the bee to its weight is (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]. To hang suspended in the air without effort, the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).
Explanation:
(a) How many electrons were added to or removed from the honeybee?
To determine the number of electrons added to or removed from the honeybee, we can use the formula:
q = ne
Where q is the charge in Coulombs, n is the number of electrons, and e is the elementary charge, which is the charge of a single electron.
Given that the charge acquired by the honeybee is +24 pC (pC = picocoulombs = 10⁻¹² C), we can substitute the values into the formula:
24 x 10⁻¹² C = n x (1.6 x 10⁻¹⁹ C)
Solving for n, we find:
n = 15,000 electrons
Therefore, the honeybee gained 15,000 electrons.
(b) What is the ratio of the electric force on the bee to the bee’s weight (Fe/Fg)?
To calculate the ratio of the electric force on the bee to its weight, we can use the formula:
F = Eq
Where F is the force, E is the electric field strength, and q is the charge.
Given that the electric field near the surface of the Earth is 100 N/C downward, we can substitute the values into the formula:
F = (100 N/C) x (24 x 10⁻¹² C)
Using the weight formula Fg = mg, where m is the mass and g is the acceleration due to gravity, we can find the weight of the bee.
Fg = (0.12 g) x (9.8 m/s²)
Taking the ratio of Fe to Fg, we get:
Fe/Fg = (100 N/C) x (24 x 10⁻¹² C) / [(0.12 g) x (9.8 m/s²)]
(c) What electric field strength and direction would allow the bee to hang suspended in the air without effort?
The electric field needed for the bee to hang suspended in the air without any effort would need to balance the gravitational force acting on the bee. The gravitational force on the bee can be calculated using the formula Fg = mg, where m is the mass of the bee and g is the acceleration due to gravity. We can then set this force equal to the electric force using the formula F = Eq. Rearranging the formula, we get E = F/q.
Since we want the bee to hang suspended without any effort, the electric force needs to be equal in magnitude and opposite in direction to the gravitational force. This means the electric field strength should have the same magnitude as the gravitational field strength (100 N/C downward), but in the opposite direction (upward).
What is the resolution of an analog-to-digital converter with a word length of 12 bits and an analogue signal input range of 100V? Show work.
The resolution of an ADC with a word length of 12 bits and an input range of 100V is approximately 0.0244V. Calculating this gives us a resolution of approximately 0.0244V.
Explanation:The resolution of an analog-to-digital converter (ADC) is determined by the number of bits used to represent the digital output.
In this case, the ADC has a word length of 12 bits.
The resolution can be calculated using the formula:
Resolution = Full Scale Range / (2^Word Length)
In this case, the Full Scale Range is 100V. Plugging in the values:
Resolution = 100V / (2^12)
Calculating this gives us a resolution of approximately 0.0244V.
sky divers jump out of planes at an altitude of 4000m. How much timewill passuntill they deploy their parachutes at an altitutde of 760m?
Answer:
Explanation:
Given
Altitude from which Diver steps out [tex]H=4000\ m[/tex]
altitude at which diver opens the Parachutes [tex]h=760\ m[/tex]
total height covered by diver[tex]=H-h=4000-760=3240\ m[/tex]
Initial velocity of diver is zero
using equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where s=displacement
u=initial velocity
a=acceleration
t=time
[tex]3240=0+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=661.224[/tex]
[tex]t=25.71\ s[/tex]
A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0°, and 60.0°.
The height and distance for 30 degrees are 14.4 m and 100 m, for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.
In projectile motion, consider the motion along vertical and horizontal separately.
Given:
Initial velocity, [tex]u =33.6\ m/s\\[/tex]
For angle [tex]30^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(30.0)\\=29.14m/s\\u_y =33.6\times sin30.0\\ =16.80m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{16.80^2}{2\times9.8}\\=14.4\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.14}{9.8}\\=100\ m[/tex]
For angle [tex]45^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(45.0)\\=23.75m/s\\u_y =33.6\times sin45\\ =23.75m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{23.75^2}{2\times9.8}\\=28.77\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times23.75\times23.75}{9.8}\\=115.11\ m[/tex]
For angle [tex]60^o[/tex], the initial and final velocities are calculated as:
[tex]u_x =33.6\times cos(60)\\=16.8m/s\\u_y =33.6\times sin60\\ =29.09m/s[/tex]
The height and distance traveled are computed as:
[tex]h=\frac{u_y^2}{2g}\\=\frac{29.09^2}{2\times9.8}\\=43.17\ m\\R= \frac{u^2sin^22\theta}{g}\\= \frac{2\times16.80\times29.09}{9.8}\\=100\ m[/tex]
Therefore, the distance and height for 30 degrees are 14.4 m and 100 m for 45 degrees are 28.77 m and 115.11 m, and for 60 degrees are 43.17 m and 100 m, respectively.
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To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. The distance traveled can be calculated using the formula d = v²sin(2θ) / g.
To calculate the height and distance traveled by a baseball hit at different angles, we can use the equations of projectile motion. The height can be calculated using the formula h = (v²sin²θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s²). Once we have the height, we can calculate the distance traveled using the formula d = v²sin(2θ) / g.
Height (h) = (33.6²sin²30.0°) / (2 * 9.8) = 19.22 meters
Distance (d) = 33.6²sin(2 * 30.0°) / 9.8 = 152.19 meters
Height (h) = (33.6²sin²45.0°) / (2 * 9.8) = 38.45 meters
Distance (d) = 33.6²sin(2 * 45.0°) / 9.8 = 203.43 meters
Height (h) = (33.6²sin²60.0°) / (2 * 9.8) = 57.67 meters
Distance (d) = 33.6²sin(2 * 60.0°) / 9.8 = 228.05 meters
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A crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one end. What is the fundamental frequency if the tube is 0.203-m long, by taking air temperature to be 37.0ºC?
Answer:
435.467980296 Hz
Explanation:
T = Temperature = 37.0ºC
L = Length of the tube = 0.203 m
Speed of sound at a specific temperature is given by
[tex]v=331.4+0.6\times T\\\Rightarrow v=331.4+0.6\times 37\\\Rightarrow v=353.6\ m/s[/tex]
Frequency is given by (one end open other end closed)
[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{353.6}{4\times 0.203}\\\Rightarrow f=435.467980296\ Hz[/tex]
The fundamental frequency is 435.467980296 Hz
The fundamental frequency of a tube, considered as an approximation for the vocal apparatus and closed at one end, with a length of 0.203 meters and at an air temperature of 37°C, is calculated using the formula for frequency f=Vw/(4L). Substituting in the values, the fundamental frequency is about 432 Hz.
Explanation:The given question involves a concept from physics, specifically sound waves and resonance. When considering the breathing passages and mouth as a resonating tube closed at one end, the fundamental resonant frequency can be calculated using the formula Vw = fa, where Vw is the speed of sound, f represents frequency, and a is the wavelength.
For a tube closed at one end, the wavelength (λ) is four times the length of the tube. So, λ = 4L. We can rearrange the formula to solve for fundamental frequency: f = Vw / a = Vw / (4L).
If the air temperature is 37°C, the speed of sound (Vw) in air is approximately 351 m/s. Substituting these values in, you get: f = 351m/s / (4*0.203m) ≈ 432 Hz. So, the fundamental frequency of this tube (our vocal apparatus approximation) at 37°C would be about 432 Hz.
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A sperm whale can accelerate at about 0.120 m/s 2 0.120 m/s2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 1.10 m/s 1.10 m/s and accelerates to a speed of 2.05 m/s 2.05 m/s ? Assume the whale travels in a straight line.
Answer:
The distance traveled by the sperm whale is 12.46 meters.
Explanation:
Given that,
Acceleration of a sperm whale, [tex]a=0.12\ m/s^2[/tex]
Initial speed of the whale, u = 1.1 m/s
Final speed of the whale, v = 2.05 m/s
We need to find the distance traveled by the sperm whale. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
d is the distance covered
[tex]d=\dfrac{v^2-u^2}{2a}[/tex]
[tex]d=\dfrac{(2.05)^2-(1.1)^2}{2\times 0.12}[/tex]
d = 12.46 meters
So, the distance traveled by the sperm whale is 12.46 meters. Hence, this is the required solution.
A cylinder with a piston holds 0.50 m3 of oxygen at an absolute pressure of 4.0 atm. The piston is pulled outward, increasing the volume of the gas until the pressure drops to 1.0 atm. If the temperature stays constant, what new volume does the gas occupy?
Answer:
2 cubic meters
Explanation:
Suppose ideal gas, when the temperature stays constant, then the product of pressure and volume is the same as before
[tex]P_1V_1 = P_2V_2[/tex]
where [tex]P_1 = 4 atm, V_1 = 0.5m^3[/tex] are the pressure and volume of the gas before. [tex]P_2 = 1 atm, V_2[/tex] are the pressure and volume of the gas after. We can solve for the volume after
[tex]V_2 = \frac{P_1V_1}{P_2} = \frac{4*0.5}{1} = 2 m^3[/tex]
A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horizontal component of the velocity.
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
5. An automobile cooling system holds 16 L of water. How much heat does it absorb if its temperature rises from 20o C to 90o C?
Answer:
[tex]Q=4704000J\\Q=4.7MJ[/tex]
Explanation:
Given data
Heat capacity c=4.2 J/gC
Water weigh m=1000g
Temperature Risen T=20°C to 90°C
To find
Heat Q
Solution
The formula to find Heat is:
Q=mcΔT
Where
m is mass
c is heat capacity
ΔT is change in temperature
So
[tex]Q=4.2(16*1000)*(90-70)\\Q=4704000J\\Q=4.7MJ[/tex]
The automobile cooling system would absorb approximately 4,699,840 Joules of heat when its temperature rises from 20°C to 90°C, according to the formula for heat absorption Q = mcΔT.
Explanation:To answer this question, we apply the formula for heat absorption: Q = mcΔT.
Here, 'Q' is the total heat absorbed, 'm' is the mass of the substance (water in this case), 'c' is the specific heat capacity of the substance, and 'ΔT' is the change in temperature. The mass can be defined as the volume of water multiplied by the density of water (1 g/cm^3). So for 16L, m = 16000g. The specific heat capacity of water is approximately 4.186 J/g °C, and the change in temperature, ΔT, is 70°C (90°C - 20°C).
Plugging in these values, we get Q = 16000g× 4.186 J/g °C × 70°C = 4,699,840 Joules. Therefore, it's estimated that the automobile cooling system would absorb around 4,699,840 Joules of heat.
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a skydiver jumps out of a plane waering a suit and decleration due to air i s 0.45 m/s^2 for each 1 m/s, skydyvers initial value problems that models the skydiver velocity is v(t). temrinal speed assuming that accelration due to gravity is 9.8m/s
Answer:
[tex]v = 21.77\ m/s[/tex]
Explanation:
Given,
Air resistance, = 0.45 m/s² for each 1 m/s
Air resistance for velocity v = 0.45 v.
Terminal velocity = ?
acceleration due to gravity= g = 9.8 m/s²
now,
a = 0.45 v
we know,
[tex]a =\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=0.45 v[/tex]
[tex]\dfrac{dv}{v} = 0.45 dt[/tex]
integrating both side
[tex]\int \dfrac{dv}{v} = 0.45\int dt[/tex]
[tex] ln(v) = 0.45 t[/tex]
[tex]t = e^{0.45 t}[/tex]
When a body is moving at terminal velocity, Force due to gravity is equal to force due to air resistance.
m a = m g
0.45 v = 9.8
[tex]v = \dfrac{9.8}{0.45}[/tex]
[tex]v = 21.77\ m/s[/tex]
Hence, the terminal velocity of the skydiver is equal to 21.77 m/s.