Ask question

Ask question

All categories

2 years ago

5

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

the potential difference between the plates as they are being separated?

1 answer:

White raven [17]2 years ago

3 0

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

You might be interested in

A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas

vodka [1.7K]

Answer:

$v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

$F=qE$

$F_p=F_e$

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

$m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE$

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

$\frac{m_ev_e^2}{2d}=qE$

we can equals these expressions for both proton and electron, because the forces qE are the same:

$\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}$

4 0

2 years ago

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m 3, po

liberstina [14]

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity = -3.1 m/s, negative sign indicates it is pointing downward.

acceleration = - 3.7m/ $s^2$ . Negative means acceleration is towards center of planet Mars.

Time taken = 3 seconds

$v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s$

So jumping velocity of Julia = 8 m/s

6 0

2 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

Ferdinand's initial velocity: $V_{o}=5 m/s$

Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

$\theta=36.86\°$ This is the angle at which Ferdinand the frog jumps between lily pads

4 0

2 years ago

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has

NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, $\dfrac{m}{t}=30\ kg/s$

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0

2 years ago

A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an angle of 35°

spin [16.1K]

<em>Answer</em>

Force = 170 N

<em>Explanation</em>

First find the distance (d) travelled by the bulldozer.

Sin 35 = 15/d

d = 15/(sin 35)

= 26.15m

Now;

work done = force × distance.

4500 J = force × 26.15

dividing both sides by 26.15,

Force = 4500/26.15

= 172.07 N

Answer to two significant figures = 170 N

3 0

2 years ago

Read 2 more answers