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Anastaziya [24]
3 years ago
5

A solid silver sphere of radius a = 2.5 cm has a net charge Qin = - 3.0 mC. The sphere is surrounded by a concentric copper sphe

rical shell of inner radius b = 6.0 cm and outer radius c = 9.0 cm. The shell has a net charge Qout = +2.0 mC. A) What is the electric potential at the outer edge of the copper shell, given the potential at infinity is zero? B) What is the electric potential on the inner edge of the copper shell? C) What is the electric potential at the center of the silver sphere?
Physics
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

Part a)

tex]V = -1 \times 10^8 Volts[/tex]

Part b)

V_{inner} = V_{outer} = -1 \times 10^8 volts

Part c)

V = -7.30 \times 10^8 Volts

Explanation:

Part a)

Net charge distribution on each shell is given as

On surface of radius "a"

q_a = -3.0 mC

on radius "b"

q_b = 3 mC

on radius "c"

q_c = -1.0 mC

Now potential at the outer shell is

V = \frac{kq_c}{r_c}

V = \frac{(9\times 10^9)((-1\times 10^{-3})}{0.09}

V = -1 \times 10^8 Volts

Part b)

Since copper sphere is a conducting sphere so here it will be an equi potential surface

So the potential will remain same throughout the surface of this sphere

Now we can say

V_{inner} = V_{outer} = -1 \times 10^8 volts

Part c)

Now electric potential at inner sphere is given as

V = \frac{kq_a}{r_a} + \frac{kq_b}{r_b} + \frac{kq_c}{r_c}

V = \frac{(9\times 10^9)(-3 mC)}{0.025} + \frac{(9\times 10^9)(3 mC)}{0.06} + \frac{(9\times 10^9)(-1 mC)}{0.09}

V = -7.30 \times 10^8 Volts

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G = 10 N/kg or 10 m/s2
Irina18 [472]

Answer:

a) U_{g} = 40\,J, b) \eta = 70\,\%, c) v = 20\,\frac{m}{s}

Explanation:

a) The initial potential energy is:

U_{g} = (0.2\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (20\,m)

U_{g} = 40\,J

b) The efficiency of the bounce is:

\eta = \left(\frac{14\,m}{20\,m} \right)\times 100\,\%

\eta = 70\,\%

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

K = U_{g}

U_{g} = \frac{1}{2}\cdot m \cdot v^{2}

v = \sqrt{\frac{2\cdot U_{g}}{m} }

v = \sqrt{\frac{2\cdot (40\,J)}{0.2\,kg} }

v = 20\,\frac{m}{s}

5 0
3 years ago
Pleaseeeeeee helppppp this is mah last question... thank u (. ^ ᴗ ^. )​
stepan [7]

Answer:

44.1 m

Explanation:

Consider the displacement being x:

x = (g•t²)/2

x = (9.8•3²)/2

x = (9.8•9)/2

x = 88.2/2

x = 44.1 m

4 0
3 years ago
Read 2 more answers
Radio waves from an fm station have a frequency of 103.1 mhz. if the waves travel with a speed of 3.00 ´ 10 m/s, what is the wav
ryzh [129]
The frequency of the radio waves from the fm station is:
f=103.1 MHz = 103.1 \cdot 10^6 Hz
And the speed of the waves corresponds to the speed of light:
v=3 \cdot 10^8 m/s
Therefore, the wavelength of the radio waves can be found by using the following equation:
\lambda= \frac{v}{f}= \frac{3 \cdot 10^8 m/s}{103.1 \cdot 10^6 Hz}=2.91 m
3 0
4 years ago
If a simple machine reduces the strength of a force, what must be increased?
rodikova [14]

Answer:

Option D

Explanation:

The answer is option D or "the distance over which the force is applied." Since basic machinery is created to lessen the work for us humans reducing the strength of the force means that the distance which where the force is needed to be applied needs to be increased. Increasing the distance for the machine makes it all around more efficient. One of the most common examples of basic machinery is a rake. When using a rake, you do not need to apply as much force into it to make the rake efficient. If you increase the distance on where you are raking, then the amount you are raking increases, which means it's all-around efficiency increases.

Hope this helps.

8 0
3 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

5 0
4 years ago
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