The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo
1 answer:
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
The mass of the baseball, m = 0.15 kg
The speed at which it moves, v = 30 m/s
Time at which the baseball was in contact with the bat, t = 0.75 ms
t = 0.75/1000 s
t = 0.00075 s
The impulsive force is given by the formula:
Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
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Answer:
Explanation:
Given:
Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s
Area of cross-section = 0.50 cm²
Length of channel =0.25 cm
Now for the new channel
Area of cross-section = 0.30 cm²
Length of channel =0.10 cm
let the Solute Diffusion rate of new channel = s
now equating the diffusion rate per unit volume for both the channels
thus,
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
Plss see attached file
Explanation:
