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Misha Larkins [42]
3 years ago
15

Can someone help me with the motion maps, I don't understand what I am supposed to write in the description or how to draw the a

rrows on the maps. 50 points!!!

Physics
1 answer:
Ugo [173]3 years ago
7 0

Answer:

i  d  k

Explanation:

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3 years ago
A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct
Taya2010 [7]

Answer:

VB − VA = g tAB   &   (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s

v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s

\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s

\frac{h}{t}=\frac{3.8}{0.02}=190\ s

Hence, the equations VB − VA = g tAB   &   (VA + VB)/2 = h / tAB will be used

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3 years ago
A sled is pulled up to the top of a hill. At thetop of the hill the sled is released from rest and allowed to coastdown the hill
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Answer:

a.) the speed at the bottom is greater for the steeperhill

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since the energy at the bottom of the steeper hilis greater

mgh =\frac{1}{mv^2}

As we can see from above that v is higher when h ishigher.

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