Correct answers A583N B128miles A baseball mass 145 gm surf
Solution
mass m = 145 gm
intial vel = 0
final vel. = 90 miles/hr = 90*1.61/3.6 = 40.25 m/s
duration = 10 ms
accelration = 40.25/10e-3 = 4.025 e+3 m/s2
Force = ma = 145 e-3 * 4.025e+3 = 583.625 N
need to be applied to accelerate the ball
heat transfer co-efficent h = 20 w/m2 –k
ball temp = 750 F = 23.890 C = 296.89 K
ambient temp = 1000 F = 37.780 C = 310.78 K
surface area A = 28.26in2 = 0.0182 sq.m
heat transferred q = hA(Ta – Tb )
= 20*0.0182*(310.78-296.89)
= 5.05596 J/s
specific heat of the ball = 1.5kJ /kg-k
mass of the ball = 145 gms
heat required to raise the temperature to 900 F
= 0.145*1.5e+3*(15/1.8) = 1812 J
duration of heating = 1812/5.05 = 358 s
distance travled s = at2/2 = 4.025 e+3 *3582 /2
= 2.59e+8 m
= 2.59e+5 km
= 1.62 e+5 miles
The answer does not match with the expected answer. looks like something wrong with the given data like specific heat or heat transfer co-efficient or surface area.

