The weight of air resting on a surface, divided by the area
of the surface ... described in units of force per unit area ...
is called air pressure.
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
I am a competitive figure skater. There are certain turns you can use such as a mowhawk, where you set one foot down that is facing the opposite direction from which you are gliding. There is a two foot turn, where you either go on or toes and turn backwards, or lean somewhat on your heals and turn forwards. Use your hips to help turn. And a 3 turn, which is basically a 2 foot turn on 1 foot. But remember, it takes practice, and you may fall a couple times.
Answer:
The value is
Explanation:
From the question we are told that
The amount of power delivered is 
The time taken is 
The wavelength is 
Generally the energy delivered is mathematically represented as

Where
is the Planck's constant with value 
c is the speed of light with value 
So

=> 
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