Answer:
25/30 = 5/6 m/s^2 5/6 meters per second squared
<span>The flight controls must be held with left aileron up and elevator neutral while taxiing a tricycle-gear equipped airplane with a left quartering tailwind. In aircraft, ailerons are placed on the trailing edge of each wing near the wingtips and can be moved up and down. So when the left aileron is up, the movement of the airplane moves to the left and turns the wheel in a counterclockwise direction while at the same time, the right aileron is down.
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Answer:
F = 36 kN
Explanation:
It is given that,
The pressure on the base of the fish tank is 4000N/m².
The base of the tank is a rectangle measuring 2.0m by 4.5m.
Area of the base of the tank is 9 m²
We need to find the force on the base caused by the base of the water. Pressure on the base of the tank is given by the force acting per unit area such that,
So, the force of 36 kN is acting on the base by the base of water.
Answer:
The work done on the sled by friction, W = - 4593.75 J
Explanation:
Given data,
The combined mass of sled and the boy, m = 75 kg
The displacement of the boy, S = 25 m
The coefficient of the friction, u = 0.25
The frictional force acting on the boy,
<em>F = u η</em>
Where,
η - is the normal force acting on the boy (mg)
Substituting the values,
F = 0.25 x 75 x 9.8
= 183.75 N
Since the direction of the frictional force is against the direction of motion
F = - 183.75 N
The work done on the sled by friction,
W = F x S
= - 183.75 x 25
= - 4593.75 J
Hence, the work done on the sled by friction, W = - 4593.75 J
