Respuesta :
Answer:
38.67 N
Explanation:
v = Final velocity = 40 m/s
u = Initial velocity
m = Mass of ball = 0.145kg
s = Displacement of ball = 3 m
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{40^2-0^2}{2\times 3}=\frac{800}{3}\ m/s^2[/tex]
F=ma
[tex]\\\Rightarrow F=0.145\times \frac{800}{3}\\\Rightarrow F=38.67\ N[/tex]
∴ Average force that a baseball pitcher's hand exerts on the ball is 38.67 N
Answer:
Force, F = 38.66 N
Explanation:
It is given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 0
Final speed of the baseball, v = 40 m/s
Distance covered, d = 3 m
Let a is the acceleration of the baseball. It can be calculated using the third equation of motion as :
[tex]a=\dfrac{v^2-u^2}{2d}[/tex]
[tex]a=\dfrac{(40)^2}{2\times 3}[/tex]
[tex]a=266.66\ m/s^2[/tex]
Let F is the average force that a baseball pitcher's hand exerts on a baseball. Using the second equation of motion to find it :
[tex]F=m\times a[/tex]
[tex]F=0.145\ kg\times 266.66\ m/s^2[/tex]
F = 38.66 N
So, the average force a baseball pitcher's hand exerts on a baseball is 38.66 N. Hence, this is the required solution.
