Ajwa asked in Science & MathematicsPhysics · 1 decade ago

The K.E. energy of a body is ;k; when moving with velocity v , if the linear momentum of sme body is tripled .?

the K,E. energy of a body is ;k; when moving with velocity V.if the linear momentum of the same body is tripled then the percentage change in K.E is

a)800 incerase

)900 increase

c)800 decrease

d)900 decrease

q.no 2

the K.E of mass m moving with velocity V is K.if the linear momentum is increased by 50 percent then the percentage change in KE is roughly equal to

a)15.6

b)30

c)40

d)none

q.no 3

A 1000 W bulb operates 10 hr daily the energy dissipated by the bulb is

a)100 kw

b)10 kw.hr

c)1000 kw.hr

1 Answer

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  • 1 decade ago
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    the K,E. energy of a body is ;k; when moving with velocity V.if the linear momentum of the same body is tripled then the percentage change in K.E is

    k = (1/2)mv^2 = (1/2m)*(m^2)*(v^2) = (1/2m)*(mv)^2 = (p^2)2m

    if p is tripled, k will become nine times. So there would be 800 percent increase. So response

    a)800 (percent) increrase is correct

    q.no 2

    the K.E of mass m moving with velocity V is K.if the linear momentum is increased by 50 percent then the percentage change in KE is roughly equal to

    P increases by 50 % means 100 becomes 150, meaning p becomes 1.5 times, so k becomes 1.5^2 = 2.25 times so there is 125 % increase. So response

    d)none is correct

    q.no 3

    A 1000 W bulb operates 10 hr daily the energy dissipated by the bulb is equal to 1 kw for 10 hour. So

    b)10 kw.hr is correct

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