If the architectural plans show the rough opening of a window to be 3'-3" x 4'-9" , the height of the opening should actually me
1 answer:
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The work done on the ship is
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and of the displacement
In this problem, we have:
(force acting on the ship)
d = 3.00 km = 3000 m (displacement of the ship)
(because the force is horizontal, and the displacement is horizontal as well)
Therefore, the work done on the ship is
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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