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In-s [12.5K]
3 years ago
7

Can someone pleaseeee answer this !!!!!!

Physics
1 answer:
LenaWriter [7]3 years ago
3 0

Answer:

The person with locked legs will experience greater impact force.

Explanation:

Let the two persons be of nearly equal mass (say m)

The final velocity of an object (person) dropped from a height H (here 2 meters) is given by,

v=\sqrt{2gH}

(g = acceleration due to gravity)

which can be derived from Newton's equation of motion,

v^2=u^2+2aS

Now, the time taken (say t ) for the momentum ( mv ) to change to zero will be more in the case of the person who bends his legs on impact than who keeps his legs locked.

We know that,

Force=\frac{\Delta(mv)}{t}

Naturally, the person who bends his legs will experience lesser force since t is larger.

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3 years ago
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A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the
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Answer:

 ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round  = 144 Kg

angle =  33.2°

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I = \dfrac{1}{2}\times 144 \times 2.75^2

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now ,

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d = 0.579 rad

we know,

using equation of rotational motion

d = \omega t + \dfrac{1}{2}\alpha t^2

0.579 = \dfrac{1}{2}\times 0.190\times t^2

 t = 2.46 s

angular speed

 ω =  α  x t

 ω = 0.19 x 2.46

 ω = 0.467 rad/s

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