a feather is dropped on the moon from a height of 1.40meters. the acceleration of gravity on the moon is 1.67m/s^2. determine th
1 answer:
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Answer:
e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.
Explanation:
This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.
The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.
<u>Mathematical expression for the Newton's second law of motion is given as:</u>
............................................(1)
where:
dp = change in momentum
dt = time taken to change the momentum
We know, momentum:
Now, equation (1) becomes:
<em>∵mass is constant at speeds v << c (speed of light)</em>
and,
where: a = acceleration
also
so, more the time, lesser the force.
<em>& </em><u><em>Impulse:</em></u>
∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.
So, the impulse in both the cases will be same.
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
- ∑Fₓ = maₓ
- F_f - F_g sinΘ = maₓ
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
- T₁ = m₂(a + g)
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
