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2015-02-27T19:44:12+05:30

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a ) law of conservation of energy is change one form to another form there is no change ( gain or loss ) in total energy & isolated system remain constant .
     energy before transmission = energy after transmission .
proof , let us consider a body of mass 'm' kg lies at position A at height ' h ' from the ground . at A
        P.E. = mgh 
         K .E. = 0 
       T.E. = P.E.+K.E.= mgh + 0 = mgh 
now, allow to free fall with velocity ' v ' . let it reaches the position B & covers a vertical distance x . at B
   P.E.= mg(h-x)
   K.E.= 1/2mv^2
   v^2 = u^2 + 2gx
     v^2 = 2gx
T.E. at B = mg( h-x) + 1/2m*2gm
              = mgh - mgh + mgh 
              = mgh
before striking the ground . at C 
     K.E = 1/2mv^2
         v^2 = u^2+2gh
         v^2 = 2gh 
 K.E. = 1/2m*2gh = mgh ( P.E)       ( PROVED )

b ) given that , 
mass ( m ) = 1.5 tonnes = 1500 kg 
initial velocity (u) = 36 km/h = 36 *1000/36000 = 10m/s
final velocity ( v ) = 72 m/h = 72*1000/3600 = 20m/s
time ( t) = 20 seconds 
power ( p ) = ?
work done = ?
work done = change in k.e.
=  \frac{1}{2} mv^2 - \frac{1}{2} mu^2
=  \frac{1}{2} *1500 * 20^2 - \frac{1}{2} *1500 * 10^2
= 225000 j
p = w/t = 225000/20 = 11250 w 

                      



               


1 5 1
b ) given that ,
mass ( m ) = 1.5 tonnes = 1500 kg
initial velocity (u) = 36 km/h = 36 *1000/36000 = 10m/s
final velocity ( v ) = 72 m/h = 72*1000/3600 = 20m/s
time ( t) = 20 seconds
power ( p ) = ?
work done = ?
work done = change in k.e.
= [tex] \frac{1}{2} mv^2 - \frac{1}{2} mu^2[/tex]
= [tex] \frac{1}{2} *1500 * 20^2 - \frac{1}{2} *1500 * 10^2[/tex]
= 225000 j
p = w/t = 225000/20
= 1125w
answer is 11250 w