3 years ago

Sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a

Physics

1 answer:

anygoal [31]3 years ago

8 0

Answer:

33 m/s

first i do 27+6=33 +s2=

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Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

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$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

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