To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,
Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity
PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,
From the speed it is possible to use find the formula, so
Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.
PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is
Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
Answer:
1.05m or 105cm
Explanation:
Using the hooke's law equation as follows;
F = –k.x
Where;
F = force (N)
x = extension length (m)
k = constant of proportionality (N/m)
According to the information given in this question;
Displacement (x) = 85cm = 85/100 = 0.85m
Force = 12500N
Using F = kx, we find the proportionality constant
k = F/x
K = 12500/0.85
K = 14705.8N/m.
Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;
F = kx
x = F/k
x = 15500/14705.8
x = 1.05m or 105cm
Surveys are considered the most reliable way to gather data
Answer:
B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2
C) V = 0.84m/s T = 29.92s
D) ω2 = 0.315 rad/s
Explanation:
The moment of inertia when they are standing on the edge:
where M is the mass of the merry-go-round.
I1 = 1680 kg.m^2
The moment of inertia when they are standing half way to the center:
I2 = 1120 kg.m^2
The tangencial velocity is given by:
V = ω1*R = 0.84m/s
Period of rotation:
T = 2π / ω1 = 29.92s
Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:
I1*ω1 = I2*ω2 Solving for ω2:
ω2 = I1*ω1 / I2 = 0.315 rad/s
