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amid [387]
3 years ago
10

A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was

Physics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

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The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of
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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is  =4.51m/s  

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

        Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

  Now to obtain the initial energy stored

               Let U denote the initial energy stored and

                      U  = \frac{1}{2} kx^2

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   Now referring to the formula above

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                \frac{1}{2} mv^2 = 112.86 - \mu_kmgx

                v^2 = \frac{2(112.86 -\mu_kmgx)}{m}

                v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}

and we are told that coefficient of friction  = 0.4 and the mass is 9 kg ,the acceleration due to gravity = 9.8m/s^2  this displacement length of spring = 0.6

  Therefore   v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }

                        =4.51m/s      

           

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