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levacccp [35]
3 years ago
9

A tennis ball is hit into the air with a racket. When is the ball’s kinetic energy the greatest? Ignore air resistance. when it

leaves the racket just before it reaches the ground when it reaches its maximum height when its kinetic energy and potential energy are equal just before it starts falling downward.
Physics
2 answers:
Snowcat [4.5K]3 years ago
6 0

Answer:

just before the ball reaches the ground

natita [175]3 years ago
5 0
<h3><u>Answer;</u></h3>

just before it reaches the ground

<h3><u>Explanation;</u></h3>
  • Kinetic energy is the energy possessed by a body or an object in motion.
  • <em><u>Kinetic energy is given by 1/2mv², where m is the mass of the object and V is the velocity of the body. Thus, kinetic energy depends on the velocity of the body if mass is kept constant.</u></em>
  • <em><u>As soon as the ball leaves the racket it has more kinetic energy and zero potential energy. As it moves up its velocity decreases, and thus the kinetic energy is being converted to kinetic energy up to maximum height reached where kinetic energy will be zero since the velocity is zero.</u></em>
  • <em><u>When the ball is going down the potential energy will be converted to kinetic energy up to a point just before it hits the ground, where kinetic energy is maximum since the velocity of the ball is maximum, due to gravitational acceleration.</u></em>
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A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
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Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

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An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
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b)43.5 rad/s

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e)84 kW

Explanation:

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I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

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P_i = T*\omega = 1930*43.5=83983 W or 84 kW

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3 years ago
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