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4 years ago

Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.

2 answers:

7 0

I think, the skydiver has maximum potential energy before jumping. However, as the skydiver jumps and decreases in height, he increases the velocity. Thus, as soon as the skydiver has jumped he is experiencing kinetic energy. The velocity of the skydiver increases as he moves down wards due to gravitational acceleration until he attains terminal velocity.

icang [17]4 years ago

4 0

Sample Response: Before jumping from the plane, the skydiver has potential energy. When the skydiver jumps, the potential energy is transformed into kinetic energy, which increases until the skydiver reaches terminal velocity. Potential energy is then transformed into thermal energy.

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Cellular telephones use what type of wave?

lana [24]

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3 years ago

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A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee

Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

$s = ut +\frac{1}{2}at^{2}$

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

$s = \frac{1}{2}gt'^{2}$

$s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}$ (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

$s = v\times t''$

$s = 343\times t''$ (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t' (3)

Using eqn (2) and (3):

s = 343(3.5 - t') (4)

from eqn (1) and (4):

$4.9t'^{2} = 343(3.5 - t')$

$4.9t'^{2} + 343t' - 1200.5 = 0$

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

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3 years ago

The baseball player is sliding towards the left of the picture to get to second base. In what direction is the force of friction

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3 years ago

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A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

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3 years ago