During a war between Ra1 and G1, the power shot shooted by G1 reaches to Ra1. It was found that the power of the power shot fired depends on mass ( m 0 ) of G1, velocity ( v 0 ) of his hand and ti

Question

During a war between Ra1 and G1, the power shot shooted by G1 reaches to Ra1. It was found that the power of the power shot fired depends on mass $(m_{0})$ of G1, velocity $(v_{0})$ of his hand and time lag $(t_{0})$ between his thought of firing and actual time when shot was fired. If power of power shot depends on $k^{th}$ power of velocity of his hand, fill k .

Answer:

2

Solution:

$p = k m_{0}^{a} v_{0}^{b} t_{0}^{c}$

$M L^{2} T^{- 3} = M^{a} L^{b} T^{- b + c}$

So $a = 1, b = 2$ and $- b + c = - 3 \Rightarrow c = - 1$ .

Power varies as the 2nd power of velocity. Answer: 2

TestHub · Accepted Answer

p = k m 0 a ​ v 0 b ​ t 0 c ​ M L 2 T − 3 = M a L b T − b + c So a = 1 , b = 2 and − b + c = − 3 ⇒ c = − 1 . Power varies as the 2nd power of velocity. Answer: 2

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