Answer:
c. 1600J
Explanation:
The loss in potential energy of the boy is given by:

where
m = 40 kg is the mass of the boy
g = 9.8 m/s^2 is the acceleration of gravity
is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)
Substituting, we find

Answer:
A) the maximum acceleration the boulder can have and still get out of the quarry
B) how long does it take to be lifted out at maximum acceleration if it started from rest
Explanation:
A)
let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.
the weight of the chain is:
and maximum tension is 
total mass and weight is :


∑



B)
maximum acceleration

using 
to solve for t


lovely question hope 7 solve it
Explanation:
Answer:
Satellite D has a mass (kg) of 500 and the distance from Earth (km) is 320.
Explanation:
The universal law of gravitation states that the force between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
We have to choose the satellite having greatest gravitational force with earth. In all options the distance from the earth is same i.e. 320 km. So, we have to select the satellite having maximum mass because the mass of the earth is constant.
Hence, the correct option is (D) " Satellite D has a mass (kg) of 500 and the distance from Earth (km) is 320 ".
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m