Answer:
a) m = 69.0 kg
b) release some gas in the opposite direction to the astronaut's movement
Explanation:
a) Let's use Newton's second law
F = m a
m = F / a
m = 60.0 / 0.870
m = 69.0 kg
b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved
p₀ = p_f
m v = M v '
v ’=
so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.
One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.
Answer:
A) 138.8g
B)73.97 cm/s
Explanation:
K = 15.5 Kn/m
A = 7 cm
N = 37 oscillations
tn = 20 seconds
A) In harmonic motion, we know that;
ω² = k/m and m = k/ω²
Also, angular frequency (ω) = 2π/T
Now, T is the time it takes to complete one oscillation.
So from the question, we can calculate T as;
T = 22/37.
Thus ;
ω = 2π/(22/37) = 10.5672
So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g
B) In simple harmonic motion, velocity is given as;
v(t) = vmax Sin (ωt + Φ)
It is from the derivative of;
v(t) = -Aω Sin (ωt + Φ)
So comparing the two equations of v(t), we can see that ;
vmax = Aω
Vmax = 7 x 10.5672 = 73.97 cm/s
Answer:
D) True. the protostar rotates more quickly.
Explanation:
If the system is isolated, the angular momentum must be retained.
Initial
L₀ = I w₀
Final
=
L₀ = 
I w₀ = 
= I /
w₀
In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase
Let's examine the answers
A) False. The opposite happens
B) False. Speed changes
C) False. For this there must be an external force, which does not exist
D) True. You agree with the above
Answer:
The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
Explanation:
Given;
velocity of wave on the string with lower tension, v₁ = 35.2 m/s
the fundamental frequency of the string, F₁ = 258 Hz
<u>velocity of wave on the string with greater tension;</u>

where;
v₁ is the velocity of wave on the string with lower tension
T₁ is tension on the string
μ is mass per unit length

Where;
T₁ lower tension
T₂ greater tension
v₁ velocity of wave in string with lower tension
v₂ velocity of wave in string with greater tension
From the given question;
T₂ = 1.1 T₁

<u>Fundamental frequency of wave on the string with greater tension;</u>
<u />
<u />
Beat frequency = F₂ - F₁
= 270.6 - 258
= 12.6 Hz
Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz