Answer:
A baseball (m= 149g) approaches a bat horizontally at a speed of 40.2 m/s (90 mi/h) and is hit straight back at a speed of 45.6m/s (102mi/h). If the ball is in contact with the bat for a time of 1.10ms, what is the average force exerted on the ball by the bat ? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.
Explanation:
Use the impulse equation (a form of Newton's 2nd Law): FΔt = Δ(mv) where Δ means "change in"
The change in momentum is mBB(vf - vi) = (.150 kg)(-46.9 m/s - 40.5 m/s)
Divide this by the time interval and you get F exerted by the bat in Newtons.
Take care.
Which picture are you talking about?
Answer:
Explanation:
Your question did not get pasted correctly. Please take a photo of the question and add to the original post. I will correct my answer once that is done. Thank you.
Acceleration is measured in m/s².
Answer: m/s²
Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
