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ANTONII [103]
3 years ago
15

a child sleds down a steep snow-covered hill with an acceleration of 3.94 m/s^2. if her initial speed is 0.0m/s and her final sp

eed is 21.1 m/s, how long does it take for her to travel from the top of the hill to the bottom?
Physics
1 answer:
-Dominant- [34]3 years ago
3 0

The time taken is 5.36 s

Explanation:

The motion of the child is a uniformly accelerated motion, therefore we can use the following suvat equation:

v=u+at

where

a is the acceleration

u is the initial  velocity

v is the final velocity

t is the time

For the child in this problem, we have:

u = 0

v = 21.1 m/s

a=3.94 m/s^2

Substituting and solving  for t,  we find the time taken:

t=\frac{v-u}{a}=\frac{21.1-0}{3.94}=5.36 s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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We are solving for V, so with the manipulation of variables we multiply V on both sides giving us 
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now we divide D on both sides giving us
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We know our mass which is 600g and our density is 3.00 g/cm^3
so
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A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
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The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

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Re-arranging the equation,

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it is halved.

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