Answer:
55 ft/s
Explanation:
A₁ = Area of rectangular cross-section at input side = 1.5 x 11 = 16.5 ft²
A₂ = Area of rectangular cross-section at far end = 1.5 x 6 = 9 ft²
v₁ = speed of water at the input side of channel = 30 ft/s
v₂ = speed of water at the input side of channel = ?
Using equation of continuity
A₁ v₁ = A₂ v₂
(16.5) (30) = (9) v₂
v₂ = 55 ft/s
1cathode rays plural : the high-speed electrons emitted in a stream from the heated cathode of a vacuum tube
2: a stream of electrons emitted from the cathode of a vacuum tube —usually used in plural
Answer:
Whether the force exerted by the locomotive on the wall was larger
Than the force the locomotive could exert on the wall.
Explanation:
The Newton's third law of motion States that every force have it's equal and opposite reaction force, whose magnitude is the same as the applied force. Therefore the magnitude of these opposite forces will be equal.
So we have;
F12=-F21
F12 is the force in a direction
-F21 is the force in the opposite direction.
Therefore we see that the magnitude of the force the locomotive exerts on the wall is equal to the one the wall exerts on the locomotive. Both magnitudes are equal but in opposite directions.
The magnitude of the average force exerted on the ball by the wall is calculated below.
The average force exerted by the ball on the wall is 3 N
Explanation:
Given:
mass of the ball (m)=0.10 kg
speed (v) =3.0 m/s
time taken(t) =0.01 seconds
To calculate:
Average force(F) exerted by ball on the wall
We know;
F=(m×v)÷t
F=(0.10×3.0)÷0.01
<u><em>F=3 N</em></u>
Therefore the average force exerted by the ball on the wall is 3 N
It goes in the downward direction
