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Ivanshal [37]
4 years ago
9

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

Physics
2 answers:
GaryK [48]4 years ago
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:<span> 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.</span>
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Answer:

4

Explanation:

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List two of the number of strategies you have learned
jolli1 [7]

Answer:

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Explanation:

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2 years ago
A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.
dedylja [7]

Answer:

F = 768 N                  

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, \dfrac{dm}{dt}=240\ kg/min = 4\ kg/s

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

F=\dfrac{dp}{dt}

F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}

Since, the speed is constant,

F=m\dfrac{dv}{dt}

F=v\dfrac{dm}{dt}

F=3.2\times 240

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0
4 years ago
A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of
Rudiy27

Answer:

Explanation:

Given

mass of spring m=100\ gm

extension in spring x=5\ cm

downward velocity v=70\ cm/s

Position in undamped free vibration is given by

u(t)=A\cos \omega _0t+B\sin \omega _0t

where \omega _0^2=\frac{k}{m}

also \frac{k}{m}=\frac{g}{L}

\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}

\omega _0=14

u(t)=A\cos(14t)+B\sin(14t)

it is given

u(0)=0

u'(0)=70\ cm/s

substituting values we get

A=0

u(t)=B\sin (14t)

u'(t)=14B\cos (14t)

70=14B

B=\frac{10}{2}

B=5

u(t)=5\sin (14t)

3 0
3 years ago
Based on molecular orbital theory, the only molecule in the list below that has unpaired electrons is ________.
xenn [34]
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6 0
2 years ago
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