Momentum (p) = mass × velocity
so, 480×40 = 19,200 kg km/hr
so the answer is C !!
Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0
momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ
Solve the two equations for v₂ and θ.
Answer:
a) W_total = 8240 J
, b) W₁ / W₂ = 1.1
Explanation:
In this exercise you are asked to calculate the work that is defined by
W = F. dy
As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.
W = F dy = F Δy
let's apply this formula to our case
a) Let's use Newton's second law to calculate the force in the first y = 5 m
F - W = m a
W = mg
F = m (a + g)
F = 80 (1 + 9.8)
F = 864 N
The work of this force we will call it W1
We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)
F₂ - W = 0
F₂ = W
F₂ = 80 9.8
F₂ = 784 N
The work of this fura we will call them W2
The total work is
W_total = W₁ + W₂
W_total = (F + F₂) y
W_total = (864 + 784) 5
W_total = 8240 J
b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use
W₁ / W₂ = F y / F₂ y
W₁ / W₂ = 864/784
W₁ / W₂ = 1.1
<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
Answer:
Ft
Explanation:
We are given that
Initial velocity=u=0
We have to find the magnitude of p of the momentum of the particle at time t.
Let mass of particle=m
Applied force=F
Acceleration, 
Final velocity , 
Substitute the values
We know that
Momentum, p=mv
Using the formula
