First calculate the time it would take for the crate to
fall using the formula:
h = v0 t + 0.5 g t^2
110 m = 0 + 0.5 (9.8 m/s^2) t^2
t = 4.74 s
The crate is also moving at 46 m/s on with respect to the
horizontal surface, therefore distance covered is:
d = (46 m/s) * 4.74 s
d = 217.95 m
The crate would fall 217.95 m from the tail of the car.
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
Answer:
Explanation:
Since the force applied is parallel to the displacement of the car, the work done on the car is simply given by:
where
F = 1210 N is the force applied on the car
d = 201 m is the displacement of the car
Substituting numbers into the equation, we find:
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
It is dangerous for children. Air bags are dangerous to children age 12 and under because the bag inflates at speeds up to 200 mph and that sudden blast of energy can severely injure or kill passengers who are too close to the air bag. If possible, children should ride in the center of back seat, properly restrained with a seat belt.
