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An engine pumps water continuously through a hose. Water leaves the hose with velocity v and m is mass per unit length of the water jet. If this jet hits a surface and comes to rest instantaneously, the force on the surface is : (1) mv^2 (2) (1/2)mv^2 (3) mv3 (4) (1/2) mv^3

An engine pumps water continuously through a hose. Water leaves the hose with velocity v and m is mass per unit length of the water jet. If this jet hits a surface and comes to rest instantaneously, the force on the surface is : (1) mv^2 (2) (1/2)mv^2 (3) mv3 (4) (1/2)mv^3

Grade:12

3 Answers

prashnt mishra
34 Points
7 years ago
P=dw/dtP=d(kE)/dtP=d(1/2mv^2)/dtP=1/2[d(mv^2)/dt]P=1/2[dmv^2/dt+m*2v*dv/dt]P=1/2mv^2*vP=1/2mv^3... -thnx.(p.m.)
Rockstar
13 Points
5 years ago
Rate of K. E = half mv^2÷time
We know time =distance÷speed(i)
Since it is said unit length in the ques, we take length as 1m and thus from (i) we get time =1÷v
Thus ans is (1/2mv^2)÷1/v
Or 1/2mv^3
Mohd Faiz
13 Points
5 years ago
Let m is the mass per unit length = mx/t= mv
So rate of kinetic energy is = 1/2 mv2 = 1/2 (mv)v2=mv3/2

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