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shusha [124]
3 years ago
8

The work done to lift a 10kg object to a height of 1m above the ground is

Physics
1 answer:
vlada-n [284]3 years ago
8 0
A = mg∆h = 10•10•1 = 100J
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True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the
inn [45]

Answer:

An external force is a force that acts on an object within the system from outside the system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force

Explanation:

ig the answer is true

5 0
3 years ago
The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r
Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity.  Therefore:

marble > manhole cover > basketball > wedding ring

7 0
3 years ago
If we put negative charge between two similar positive charges then what is it's equilibrium? And how?​
Gnesinka [82]

Your question has been heard loud and clear.

Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.

Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.

Thank you

5 0
3 years ago
Read 2 more answers
Please help
Allisa [31]

<u>We are given:</u>

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

<u>Net upward force on the rocket:</u>

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

<u>Upward Acceleration of the Rocket:</u>

From newton's second law:

F = ma

<em>replacing the variables</em>

300 = 10 * a

a = 30 m/s²

5 0
3 years ago
A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
2 years ago
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