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ludmilkaskok [199]
3 years ago
9

A catapult’s lever holds a cannonball. The lever is attached to a tightly held rope. When the rope is released, the lever spring

s forward and launches the cannonball. When the rope is held tightly, which form of energy does it possess?
Physics
2 answers:
MariettaO [177]3 years ago
7 0

Answer:

Explanation:

When the rope is held tightly, the rope has elastic potential energy, due to which the spring gets compressed. As it is released, the elastic potential energy of the spring is converted into the kinetic energy and the cannon balls moves forward.

Blababa [14]3 years ago
4 0

A rope held tightly posses elastic potential energy. when the rope is held tightly, it gets stretched by some amount and in a way the rope gets deformed. The rope tends to get its original shape back when the rope is released. some work is done to deform or stretch the rope against the restoring force between the particles and this work done reflects as elastic potential energy.

Elastic potential energy.

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3 years ago
Vector sum of 15 m, East and 9 m, West is 24 m, East. True False
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False

Explanation:

Because when you go through east

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I hope that it's a clear ") .

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Which bar graph could represent the reaction rates of a reversible reaction<br> that has just begun?
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At what distance from a 27 mw point source of electromagnetic waves is the electric field amplitude 0. 060 v/m ?
n200080 [17]

The distance from a 27 mw point source of electromagnetic waves where the electric field amplitude 0. 060 v/m will be 21.21 m .

Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.

The highest point of a wave is known as 'crest' , whereas the lowest point is known as 'trough'. Electromagnetic waves can be split into a range of frequencies. This is known as the electromagnetic spectrum.

c = 3 * 10^{8} m/s

∈ = 8.85 * 10^{- 12} C^{2} / N/ m^{2}

E = 0. 060 v/m

I = P / 4πr^{2}

Also , I = c ∈ E^{2} /2

r^{2} = P / 4π I                                 equation 1

substituting the value of I in equation 1

r^{2} = 2 P / 4π (c ∈ E^{2} )

r^{2} = 2 * 27 * 10^{-3} / 4 * 3.14 * 3 * 10^{8} * 8.85 * 10^{- 12}  * (0.06)^{2}

r = 21.21 m

To learn more about Electromagnetic waves  here

brainly.com/question/12392559

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