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3 years ago

Gravitational potential energy = the energy due to an objects

Physics

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

Gravitational energy is the energy of a body when it is taken to a certain height with respect to the earth surface.It is one of the part of potential energy that comes in thr part of the mechanical energy. the work done in taking a body upto a certain height get stored in the body in form of gravtitational potential energy. potential energy is directly proportional to height or altitude upto which the body is taken, mass of the body and gravtational acceleration(9.8m/s^2)So P.E. = mgh

Hope it helps

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A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free

REY [17]

Explanation:

According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely falling object.

3 0

3 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

$\theta=5.71^{o}$

8 0

3 years ago

A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?

Anna35 [415]

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2

8.5 = 4.095 t^2 Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2 Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

6 0

3 years ago

A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground. How fast is th

zysi [14]

Answer:

0.83 m/s

Explanation:

FIrst of all, we have to find the time of flight, i.e. the time the baseball needs to reach the ground. This can be done by using the equation for the vertical motion:

$h=ut+\frac{1}{2}gt^2$

where

h is the initial height

u = 0 is the initial vertical velocity

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

t is the time

Substituting h = 1.8 m and solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.8)}{9.8}}=0.61 s$

We know that the horizontal distance travelled by the ball is

d = 0.5 m

Therefore, we can find the horizontal velocity (which is constant during the whole motion):

$v= \frac{d}{t}=\frac{0.5}{0.60}=0.83 m/s$

4 0

3 years ago

What demonstrates conduction

Anastaziya [24]

Heating a pot of water on a stove top or benson burner. The water is heated through direct contact.

5 0

3 years ago