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ivann1987 [24]
3 years ago
7

Which of these is exhibiting kinetic energy? Which of these is exhibiting kinetic energy? the high-energy phosphate bonds of a m

olecule of ATP a space station orbiting Earth a rock on a mountain ledge a person sitting on a couch while watching TV an archer with a flexed bow Submit
Physics
1 answer:
Wittaler [7]3 years ago
7 0

Answer:

A space station orbiting

Explanation:

As we know that kinetic energy associated with the motion of particle and it is given as

m= Mass

V =Velocity

KE=\dfrac{1}{2}mV^2

If V= 0 ,means velocity of particle is zero then it kinetic energy will be zero.

A)

At ATP molecule does not have any movement that is why it kinetic energy is zero.

B)

A space station have velocity that is why it have kinetic energy.

C)

A rock on mountain ledge do not have kinetic energy .Because it is not moving.

D)

When person sitting it means that it dies not have any motion that is why it does not have kinetic energy.

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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

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