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andrezito [222]
3 years ago
12

A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential ene

rgy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Physics
1 answer:
Tamiku [17]3 years ago
3 0

Answer:

elastic energy

Explanation:

When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.

During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.

The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:

<u>elastic energy</u>

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3 years ago
Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST
Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

a^2+b^2=c^2

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8 0
2 years ago
Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw
Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of  the distance between them, acting along  the  line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the  factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0
3 years ago
30 POINTS!!! CAN U AWNSER IT?? :)
solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

F\Delta t = m \Delta v As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, \Delta v = 69.4 m/s;

45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg

7 0
2 years ago
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