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12345 [234]
3 years ago
14

In which one(s) of the following situations will there be an INCREASE in Kinetic Energy? Group of answer choices A block slides

down a frictionless incline A box is pulled across a rough floor at constant speed. A stone at the end of a string is whirled in a horizontal circle at constant speed. A projectile approaches its maximum height. A merry go round rotates faster due to the push by a person.
Physics
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

The Kinetic Energy INCREASE in the following situations:

a)A block slides down a frictionless incline:

e)A merry go round rotates faster due to the push by a person.

Explanation:

The energy kinetics is proportional to the square of the velocity:

E_k=1/2*mv^2

We study the cases:

a)A block slides down a frictionless incline:

The gravity made a positive work and the box get a extra velocity, then the kinetics energy INCREASE

b)A box is pulled across a rough floor at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

c)A stone at the end of a string is whirled in a horizontal circle at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

d)A projectile approaches its maximum height:

If the projectile approaches its maximum height, its velocity approaches to zero. Then the kinetics Energy DECREASE

e)A merry go round rotates faster due to the push by a person.

Thanks to the push, the magnitude of the velocity INCREASE, so the kinetics energy INCREASE as well

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Explanation:

Given the following data;

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Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

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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

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

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

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<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.

Learn more about projectiles here: brainly.com/question/11049671

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