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vazorg [7]
3 years ago
14

Two parallel plates have a potential difference between them. The distance between the plates is halved, but the electric field

between them remains constant. How did the potential difference change?
A. Potential difference was halved.


B. Potential difference did not change.


C. Potential difference was doubled.


D. Potential difference was quadrupled.
Physics
1 answer:
zhuklara [117]3 years ago
8 0

For a set of very long (essentially infinitely long) charged parallel plates, the electric field between them is constant, and the electric field is given by

E = ΔV/d

ΔV is the potential difference between the plates and d is the distance of their separation.

If d is halved but E stays constant, then by the relationship above, ΔV must also have been halved.

A. Potential difference was halved.

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An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip
OverLord2011 [107]

Answer:

The value is v_2 =  5.53 \  m /s

Explanation:

From the question we are told

  The pipe diameter at location 1 is  d  = 8.8 \  cm =  \frac{8.8 }{10} = 0.88 \ m

   The velocity at location 1 is  v_1 =  2.4 \  m /s

   The diameter at location 2 is  d_2 =  5.80 \  cm  =  0.58 \  m

Generally the area at location 1 is  

       A_1 =  \pi *  \frac{d^2}{ 2}

=>     A_1 =  \pi *  \frac{0.88^2}{ 2}

=>     A_1 = 3.142 *  \frac{0.88^2}{ 2}

=>     A_1 = 1.2166 \  m^2

Generally the area at location 1 is  

       A_2 =  \pi *  \frac{d_1^2}{ 2}

=>     A_2 =  \pi *  \frac{0.58^2}{ 2}

=>     A_2 = 0.528  \  m^2

Generally from continuity equation we have that

     A_1 * v_1 =  A_2 * v_2

=>   1.2166 *   2.4   =  0.528   * v_2

=>   1.2166 *   2.4   =  0.528   * v_2

=>    v_2 =  5.53 \  m /s

3 0
3 years ago
A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

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Answer:

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Notation

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Ff = force of the friction

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