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zaharov [31]
3 years ago
9

A 5.0 kg wooden block Is 100 meters from the ground. What is the gravitational potential energy for the block?

Physics
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

<h2>4900 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 5 × 9.8 × 100

We have the final answer as

<h3>4900 J</h3>

Hope this helps you

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Which scenario involves kinetic energy transforming into potential energy? A. a book lying on a shelf B. a train traveling at a
NeX [460]
The answer is C.

The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
6 0
3 years ago
Who has the greater velocity, an astronaut who has just completed an orbit of the Earth or you when you have just traveled from
Alik [6]

Answer:

the answer is

Explanation:

constant acceleration

because when the object's velocity is changing then the object is accelerating or decelerating

as acceleration describe changing of velocity so the answer is constant acceleration

Acceleration is defined as the rate of change of velocity.

Acceleration = (Change in velocity) / time taken

Acceleration = (Final velocity - initial velocity) / time

As the object velocity changes by the same amount in each second, it means the acceleration is constant.

Hope I can help u

4 0
2 years ago
The direction of an electric field is the direction (5 points)
vlabodo [156]

The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field  Therefore the answer is b. a positive charge will move in the field.

5 0
2 years ago
A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> I NEED THE ANSWER NOW
FrozenT [24]
90 F = 43 OR 0.9F = 0.43
(F = 43 / 90 OR 0.43 / 0.9 =) 0.48 N

upwards force = downwards force
(R =) 1.2 N
5 0
3 years ago
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