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jeka94
3 years ago
10

A 50gk boy on a rough horizontal ground the coefficeint of static friction is 0.68

Physics
1 answer:
Soloha48 [4]3 years ago
4 0

Given that,

Mass of a boy = 50 kg

The coefficient of static friction = 0.68

To find,

Let us assume we need to find the maximum static friction between the boy and the ground.

Solution,

The formula for the maximum static friction between the two objects is given by :

F=\mu N

Where

N is normal force

Substitute all the values,

F=0.68\times 50\times 10\\\\F=340\ N

Therefore, the maximum static friction between the boy and the ground is 340 N.

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now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,

K=(γ-1)E_{0,\pi }

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u'=+-0.283c

note +-=±

To find speed least and greatest speed of meson we would use relativistic velocity addition equations:

u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c

6 0
3 years ago
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