Maybe this will help you out:
Momentum is calculate by the formula:
Where:
P = momentum
m = mass
v = velocity
The SI unit:
So the unit of momentum would be:
Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:
I = FΔt
where:
I = impulse
F = Force
Δt = change in time
The SI unit:
F = Newtons (N) or 
t = Seconds (s)
So the unit of impulse would be derived this way:
I = FΔt
I = x 
or
You can then cancel out one s each from the numerator and denominator and you'll be left with:
So then:
Momentum: Impulse
First, let's put 22 km/h in m/s:
Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by
The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:
Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
Answer:
v₃ = 5 [m/s]
Explanation:
To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.
P = m*v
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision
Pbeforecollision = Paftercollision
(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)
where:
m₁ = mass of the truck = 3000 [kg]
v₁ = velocity of the truck = 10 [m/s]
m₂ = mass of the car = 1000 [kg]
v₂ = velocity of the car before the collision = 0 (the car is parked)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 15 [m/s]
(3000*10) + (1000*0) = (3000*v₃) + (1000*15)
30000 = 3000*v₃ + 15000
3000*v₃ = 30000 - 15000
3000*v₃ = 15000
v₃ = 5 [m/s]
Explanation:
Artificial gravity can be created using a centripetal force. A centripetal force directed towards the center of the turn is required for any object to move in a circular path. In the context of a rotating space station it is the normal force provided by the spacecraft's hull that acts as centripetal force.
Hope it helps.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
