Answer:
= 4.38 × 10³⁴kgm²/s
Explanation:
Given that,
mass of moon m = 9.5 × 10²²kg
Orbital radius r = 4.28 × 10⁵km
Orbital period T = 28.9days
T = 28.9 × 24 × 60 × 60
= 2,496,960s
Angular momentum of the moon about the planet
L = mvr
L = mr²w

The correct choice is
B.
Particles at the bottom of the water carry heat energy to the top of the water.
when pot of water is heater, the bottom of pot gets heated. the particles of water in contact with the bottom of the pot gets heat through conduction. after getting heat, these particles of water near the bottom, move away towards top and their position is taken by cooler particles from top. that way heat travels
Answer:
The gauge pressure of air is 110 kpa
Explanation:
Atmospheric pressure,
= 101 Kpa


where;
ρw is the density of water = 1000 kg/m³
ρo is the density of oil = 800 kg/m³
ρHg is the density of mercury = 13,600 kg/m³
g is acceleration due to gravity = 9.8 m/s²

Therefore, the gauge pressure of air is 110 kpa