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3 years ago

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What happens to the potential energy of a stationary charge when it begins to move freely from one point to another under the in

fluence of an electric field?
It increases.
It decreases.
It remains the same.
It becomes zero.

2 answers:

Ulleksa [173]3 years ago

8 0

It decreases and the decrease in the energy level appear as a kinetic energy of the charge

german3 years ago

8 0

Answer: Option (b) is the correct answer.

Explanation:

Potential energy is the energy obtained by an object due to its position. Whereas kinetic energy is the energy obtained by an object due to its motion.

For example, when a charge moves from one point to another then it means the charge has come to motion.

Thus, we can conclude that potential energy of a stationary charge when it begins to move freely from one point to another under the influence of an electric field decreases.

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the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

dem82 [27]

Answer:

<em>Time = 5 seconds</em>

<em>Distance = 50 meters</em>

Explanation:

<u>Constantly Accelerated Motion</u>

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

$v_f=v_o+a.t$

And the distance s is

$\displaystyle s=v_o.t+\frac{gt^2}{2}$

Given the object starts from rest, vo=0 and vf=20 m/s at $a=4\ m/s^2$ . We compute t

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}$

$\boxed{t=5\ sec}$

Now we compute s

$\displaystyle s=0+\frac{4\times 5^2}{2}$

$\boxed{s=50\ m}$

5 0

3 years ago

Which type of light-matter interaction causes glare?

AleksandrR [38]

Answer:

I think the reflection of light off of a shiny surface is the answer... Hope this helps

Explanation:

3 0

3 years ago

A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet

astraxan [27]

Answer:

Acceleration = 2.35 m/ $s^{2}$

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = $\frac{mg}{2}$ = 0.5 mg

Friction along incline = umg cos30° = mg $0.3\times \frac{\sqrt{3}}{2}$

Friction along incline = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = $\frac{net force}{mass}$ = 0.24 g = 2.35 m/ $s^{2}$

The height of incline = 8 m

Length of the inclined edge = 16 m

$v^{2}=u^{2}+2as$

$v^{2}= 2\times 0.24 \times 9.8\times 16$

v= 8.67 m/s

5 0

4 years ago

Do rolling down a grassy hill has kinetic energy or potential energy

Ann [662]

If you're moving, then you have kinetic energy.

If you're not at the bottom yet, then you still have
some potential energy left.

4 0

4 years ago

Read 2 more answers

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0

3 years ago