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Sergio039 [100]
3 years ago
5

A young diver is practicing his skills before an important team competition. Use the diagram below in order to analyze the energ

ies of the diver and complete the statements below.
Where m = mass (kg), g = 9.8 m/s2, v = velocity (m/s), h = height (m), KE = kinetic energy (J), and GPE = gravitational potential energy (J).
Use the equations above to answer the following questions.

A diver with a mass of 90 kg is at a height of 10 m, and he has not jumped off of the board yet (v = 0 m/s). When the diver reaches a height of 5 m (Point C), his gravitational potential energy is

A. 1350 J

B. 8820 J

C. 4410 J

D. 0 J

and his velocity is

E. 4.5 m/s

F. 0 m/s

G. 3.2 m/s

H. 9.9 m/s

Please help will mark brainliest

Physics
2 answers:
nika2105 [10]3 years ago
7 0
  • h=10m-5m=5m
  • m=90Kg
  • g=9.8m/s^2

\\ \sf\longmapsto GPE=mgh

\\ \sf\longmapsto GPE=90(5)(9.8)

\\ \sf\longmapsto G PE=4410J

Now

It's converted to kinetic energy while reaching ground.

\\ \sf\longmapsto K.E=4410

\\ \sf\longmapsto \dfrac{1}{2}mv^2=4410

\\ \sf\longmapsto 90v^2=8820

\\ \sf\longmapsto v^2=98

\\ \sf\longmapsto v=9.9m/s

Done

Lyrx [107]3 years ago
5 0

Answer:

Hope it will help you a lot.

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Astronauts brought back 500 lb of rock samples from the moon. how many kilograms did they bring back? 1 kg = 2.20 lb 227 kg 227
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<h3>Who is an astronaut?</h3>

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Explanation:

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A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat
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If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 \text{m/s}^{2}.

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is \omega=5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, a=\omega ^2 r.

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}

The acceleration is towards upwards that means towards the center of the wheel.

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