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3 years ago

A man drops a ball off of a 7.2 m high cliff, what is the ball's velocity as it hits the ground?

Physics

1 answer:

QveST [7]3 years ago

6 0

Answer:

19.3 m/s

Explanation:

Take down to be positive. Given:

Δy = 19 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)

v = 19.3 m/s

<em>My camera quality is bad. Sorry..</em>

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Which of the examples below is an example of convection?

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You were driving your car to UTD at a speed of 35 miles per hour. You stopped at the FloydCampbell intersection with the signal

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Answer:

green light have high energy

Explanation:

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The energy of the signal is given by $E=h\nu =h\frac{c}{\lambda }=\frac{6.67\times 10^{-34}\times 3\times 10^{8}}{6.45\times 10^{-7}}=3.1023\times 10^{-15}j$

The frequency of the green light is given by:

$f=5.80\times 10^{14}s^{-1}$

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So green light have high energy

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3 years ago

A ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit

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As the ball is moving in air as well as we have to neglect the friction force on it

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3 years ago

Two extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upp

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3 years ago

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

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2 years ago