Question

A baseball weighs 5.19 oz. what is the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher at 96.0 mi/h? by what factor will the kinetic energy change if the speed of the baseball is decreased to 55 mi/h

Answer 1

Kinetic energy =0.5*mas*velocity^2
Joules =lg*m^2/s^2
1 miles= 1608.34 meters
1 hour= 3600 Sec
1 ounce =28.35g =0.02836 kg
What is a the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher at 96.0 mi/h?

Answer: KE=0.5m*v^2
=0.5*(5.12 o *0.02835 kg/1 ounce)* (95 miles/h*1609.34m/1 miles* 1hr/3600s^)2
131kg*m^2/s^2= 131 joules

By what factor with the kinetic energy change if the speed of the baseball is decreased to 55.0 mi/h?

Answer: KE=0.5*m*v^2
=0.5*(5.13 o*0.02835kg/1 ounce)*(55 miles/ h*1609.34m/1 mile*1 hr/3600s)^2
=44.0kg*m^2s^2=44.0 joules

131/44= 2.98, so decreased by a factor of approximately 3

Answer 2

Answer:

Part a)

$KE = 135.3 J$

Part b)

Factor by which KE is changed is f = 0.33

Explanation:

As we know that kinetic energy of the ball is given as

$KE = \frac{1}{2} mv^2$

in order to find kinetic energy in Joule unit we need to plug in all data in SI units

so here we have

$m = 5.19 oz = 0.147 kg$

now the speed of the ball is

$v = 96.0 mi/h$

$v = 96.0 \times \frac{1609 m}{3600 s} = 42.9 m/s$

so we will have

$KE = \frac{1}{2}(0.147)(42.9)^2$

$KE = 135.3 J$

Now if the speed of the ball is 55 mph

so it is

$v = 55 \times \frac{1609}{3600}$

$v = 24.6 m/s$

so new kinetic energy is given as

$KE_2 = \frac{1}{2}(0.147)(24.6)^2$

$KE_2 = 44.4 J$

So the factor by which KE is changed is given as

$f = \frac{44.4}{135.3}$

$f = 0.33$