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stepan [7]
2 years ago
15

How much work was done on a charge of 3.0 to move it through 12v electric potential difference

Physics
2 answers:
erik [133]2 years ago
8 0

Answer:

  • Hence, the work done in moving the charge is 45 J.

<h2>( Too lazy to translate it to Spanish T_T )</h2>
Ray Of Light [21]2 years ago
8 0

Answer:

45 J

Explanation:

Potential difference (V)= 15 V

Charge (Q)= 3C

We know that

The potential difference(p.d) between two point in an electric circuit is defined as the amount of work done in moving a unit charge from one point to the other point.

Potential difference (V) = work done (W)/Charge (Q)

V= W/Q

Putting the values in the above formula we get:

15 V = W / 3 C

W= 15 V × 3 C

W= 45 J

Hence, the work done in moving the charge is 45 J.

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6 0
3 years ago
What are scientists who study space called?
Brilliant_brown [7]

Answer:

Astronomer

Explanation:

A scientist who studies the objects in the sky, including planets, galaxies, black holes, and stars, is called an astronomer. These days, the terms astronomer and astrophysicist are used interchangeably, to talk about any physicist who specializes in celestial bodies and the forces that affect them.

4 0
3 years ago
Read 2 more answers
One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
Please help me on questions 3 and 6, thank you! :D I'll give brainliest!
Anarel [89]

Answer:

3. if you increase your mass you also increase the gravitational pull

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Explanation:

i hope this helps-

4 0
3 years ago
Read 2 more answers
A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

where

\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0
3 years ago
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