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Amanda [17]
3 years ago
15

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Physics
1 answer:
Olenka [21]3 years ago
8 0

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. So we can use the following suvat equation:

v=u+at

where

v is the  final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

a=g=9.8 m/s^2 is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

v=0+(9.8)(0.75)=7.35 m/s

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 k
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Answer:

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s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2

Force is given by

F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N

The force with which the team pulls the plane is 28716.4740661 N

v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J

Work done is given by

W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J

The fraction is given by

\dfrac{54450.9540448}{106250.954045}=0.512474965841

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

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