Answer:
112m/s
Explanation:
14x8=112 therefore meaning the zebra would run 112m/s
<span>The overall force that is acting on the bottle is gravity. With the incline being 30 degrees the full force of gravity isn't acting on the bottle becuase the ramp isn't allowing the bottle to go straight down. By taking the sin of 30 degrees you find the proportion of gravity that is acting on the bottle to be 4.9 meters per second and the bottle weights 20 kg so the force acting on the bottle is 98 Newtons.</span>
Answer:
The magnitude of the average force exerted on the water by the blade is 960 N.
Explanation:
Given that,
The mass of water per second that strikes the blade is, 
Initial speed of the oncoming stream, u = 16 m/s
Final speed of the outgoing water stream, v = -16 m/s
We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :



F = -960 N
So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.
Answer:
B. τ = 16 Nm
Explanation:
In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:
τ = Fd
here,
τ = Torque = ?
F = Force exerted by the weight = Weight = mg
F = mg = (4 kg)(10 m/s²) = 40 N
d = distance from knee to weight = 40 cm = 0.4 m
Therefore,
τ = (40 N)(0.4 m)
<u>B. τ = 16 Nm</u>
Answer: The reference frame of a passenger in a seat near the center of the train
Explanation:
the speed of light is the same for the passenger and the bicyclist
then the avents are simultaneous fo the passenger not for the bicyclist
the delay between the two events for the bicyclist is
Δt=Δd/vs
where
Δd= lenght of train
vs=speed of sound
the reference frame of a passenger in a seat near the center of the train
Solution:
The space and time transformations are:
x' = γ(x - vt)
t' = γ(t - vx/c²).
In the primed frame the two events are simultaneous, so that Δt' = 0. Also here Δx' = 30. In the unprimed frame Δx' = 30 = γ(Δx - v Δt).......(*)
We also have Δt' = 0 = γ(Δt - vΔx/c²)→Δx = c²Δt/v......(**)
Substituting (**) in (*): 30 = γ(c²Δt/v - vΔt))→Δt = 30/(c²/v - v) =
30/(2c - 0.5c) = 6.7 x 10^(-8)s