Answer:
I = 0.483 kgm^2
Explanation:
To know what is the moment of inertia I of the boxer's forearm you use the following formula:
(1)
τ: torque exerted by the forearm
I: moment of inertia
α: angular acceleration = 125 rad/s^2
You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

Next, you replace this value of τ in the equation (1) and solve for I:

hence, the moment of inertia of the forearm is 0.483 kgm^2
Answer:
t = 3 [s]
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity [m/s]
Vo = initial velocity = 15 [m/s]
g = gravity acceleration = 10 [m/s²]
t = time [s]
Now replacing we have:
![0 = 15 -10*t\\10*t=15\\t= 1.5[s]](https://tex.z-dn.net/?f=0%20%3D%2015%20-10%2At%5C%5C10%2At%3D15%5C%5Ct%3D%201.5%5Bs%5D)
Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.
The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:
t = 1.5 + 1.5 = 3 [s]
<span>F = m*a = 7000kg * 9.8N/kg = 68,600 N.
68,600 N is your answer
</span>
Answer:
26.6 m/s
Explanation:
Given:
Δy = 2.1 m
t = 5.35 s
a = -9.8 m/s²
Find: v₀
Δy = v₀ t + ½ at²
(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²
v₀ = 26.6 m/s