Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = 
Where:
is relationship between exhaust velocity and specific impulse
is the porpellant fraction, also written as MR.
The relationship is: 
To determine the fraction:
Δv =
Knowing that change in velocity is Δv = 9.6km/s and 
= 9.81m/s²
<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
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<u>Part 1</u>: Isp = 450s
ln(MR) = 
ln (MR) = 2.17
MR = 
MR = 8.76
<u>Part 2:</u> Isp = 2000s
ln (MR) = 
ln (MR) = 0.49
MR = 
MR = 1.63
Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
The average current density in the wire is given by:
where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:
where r is the radius of the wire. In this problem, 
, so the cross-sectional area is
and the average current density is
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.
Since the forces are balanced the Spring force is equal to the force of the weight that is
Where,
k = Spring constant
x = Displacement
m = Mass
g = Gravitational Acceleration
Re-arrange to find the spring constant
Just before launch the compression is 40cm, then from Potential Elastic Energy definition
Therefore the energy stored in the spring is 63.72J
