Power=work/time
power=50/20
50/20=2.5
Therefore A. 2.5 W
Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.
Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)
The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)
Answer: 1.1x10¹⁶ kg
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force 
By wave particle duality.
Wavelength , λ = h / mv
where h = Planck's constant = 6.63 * 10⁻³⁴ Js, m = mass in kg, v = velocity in m/s.
m = 1kg, v = 4.5 m/s
λ = h / mv
λ = (6.63 * 10⁻³⁴) /(1*4.5)
λ ≈ 1.473 * 10⁻³⁴ m
Option D.
From the starting depth to the surface, the vertical distance is 35 ft.
From the surface to the peak of the jump, the vertical distance is 27 ft.
From the peak of the jump to the surface, the vertical distance is 27 ft.
From the surface to the ending depth, the vertical distance is 18 ft.
Then the total vertical distance is ...
35 ft + 27 ft + 27 ft + 18 ft = 107 ft
