Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction
Put the value into the formula
Hence, The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
First, find the velocity of the projectile needed to reach a height h when fired straight up.
Given:
Δy = h
v = 0
a = -g
Find: v₀
v² = v₀² + 2aΔy
(0)² = v₀² + 2(-g)(h)
v₀ = √(2gh)
Now find the height reached if the projectile is launched at a 45° angle.
Given:
v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)
v = 0
a = -g
Find: Δy
v² = v₀² + 2aΔy
(0)² = √(gh)² + 2(-g)Δy
2gΔy = gh
Δy = h/2
P1 and P2 are the pressures, and V1 and V2 are the volumes. So you take the first pressure and volume you are given and place them into the equation P1V1 so the first part of the equation would be 101000*0.5 = P2V2. You then rearrange the equation to find what you want, in this instance you would do 50500/0.25 = P2... therefore P2 = 2020000Pa or 2.02*10^6Pa
Answer:
True. mark me as Brilliant
