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trapecia [35]
3 years ago
5

An athlete is working out in the weight room.he steadily holds 50 kilograms above his head for 10 seconds which statement is tru

e about this situation
Physics
1 answer:
My name is Ann [436]3 years ago
3 0
The athlete is doing work because he prevents the weight from falling downward
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Write one general properties of non contact force​
Gekata [30.6K]

Answer:    Gravitational force between the two masses does not depend on the medium separating two masses.

3 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position
KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

W_p = KE_f - KE_i

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

KE_f = \frac{1}{2}mv_1^2

now work done is given as

W_p = \frac{1}{2}mv_1^2 - 0

so we can say

W_p = \frac{1}{2}mv_1^2

so above is the work done on the box to slide it from x1 to x2

3 0
3 years ago
Two +1 C charges are separated by 30000 m, what is the magnitude of<br> the force?
Kipish [7]

Answer:

<em>The magnitude of the force is 10 N</em>

Explanation:

<u>Coulomb's Law</u>

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:

\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}

\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}

F = 10 N

The magnitude of the force is 10 N

7 0
2 years ago
If 270 watts of power is used in 42 seconds, how much work was done<br>​
navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

W = 270(42)

<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0
3 years ago
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