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Marizza181 [45]
3 years ago
9

A skydiver jumped out of a plane, determine distance he distance he descended after 5 seconds. Gravity is pulling him down at 10

m/s^2
Physics
1 answer:
Bond [772]3 years ago
8 0

The distance travelled in 5 seconds is 125 m

Explanation:

The motion of the skydiver is a free fall motion, since he is acted upon the force of gravity only. Therefore, it is a uniformly accelerated motion, with constant acceleration downward, and we can find the distance he travels by using the following suvat equation:

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

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

In this problem, we have:

u = 0 (the skydiver jumps from rest)

a=g=10 m/s^2 (acceleration of gravity)

And substituting

t = 5.0 s

we find the distance travelled:

s=0+\frac{1}{2}(10)(5)^2=125 m

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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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
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Explanation:

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Speed of electron,v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0

Speed of second electron,v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}

v_2=1.5\times 10^8 m/s

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Explanation:

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We know that the moment of inertia of the ring is given by :

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