50POINTS! What is the escape velocity for lunar module? Lunar module mass 15,200 kg radius of moon 1.74x106m, mass of moon 7.34x
2 answers:
Answer:
2.73 km/s
Explanation:
The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)
where
is the gravitational constant
is the mass of the Moon
is the radius of the Moon
As we can see, the escape velocity does not depend on the mass of the lunar module.
Substituting the numbers into the formula, we find
You might be interested in
See the graph in attachment
Explanation:
In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.
In the graph, we see that the initial velocity at time t = 0 is
and it is negative, so below the x-axis.
Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).
After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.
Learn more about velocity:
#LearnwithBrainly
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s