As altitude increases, temperature increases.
The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>
Answer:
21870.3156 N
Explanation:
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 1.6 m/s²
Equation of motion
The acceleration of the craft should be 1.02234 m/s²
Weight of the craft
Thrust
The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
