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>
Displacement will be 15 too because
It falls from 0 till 15 meters
ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ
Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball, 
After the collision, final momentum 
Using the conservation of momentum as :
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)
