1. The stratosphere is above the troposphere. This layer of the atmosphere is where planes fly. At the top of the stratosphere, there is a ozone layer.
2. The mesosphere is above the stratosphere. Temperatures drastically drop in the mesosphere. It is the middle layer of the atmosphere.
3. Here are the layers of the atmosphere:
- Troposphere
- Stratosphere
- Mesosphere
- Thermosphere
- Exosphere
Hope this helps you!
The molecules held together are to weak so the substance melt and boil at low points.
Take gravity you are accelerated toward the ground. Since gravity has both a direction and a magnitude, then it is a vector.
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
speed Δv = v₂ = 8.15ms⁻¹
v₁ = 0ms⁻¹
time =t = 5s
acceleration = a = speed / time taken
a = Δv / Δt ( as Δv = v₂ - v₁)
a = 8.15ms⁻¹ / 5 s
a = 1.6ms⁻²
