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Bond [772]
3 years ago
8

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

1022kg
Physics
2 answers:
Airida [17]3 years ago
7 0

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)

v=\sqrt{\frac{2GM}{r}}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M=7.34\cdot 10^{22} kg is the mass of the Moon

r=1.74\cdot 10^6 m 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

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(7.34\cdot 10^{22}kg)}{(1.74\cdot 10^6 m)}}=2732 m/s=2.73 km/s

mash [69]3 years ago
6 0

Answer:

2.73 km/s

Explanation:

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A mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test
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F = 52000 N

m = 1060 kg

a= F/m = 52000 N/1060 kg = 49.0566 m/s^2

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Ms. Sparkle bought 12 cans of diet soda. Each can
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Which of the following is most closely related to an activated complex?
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Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

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3 years ago
6. Draw a velocity-time graph for an object originally traveling at -3 m/s. The object
faltersainse [42]

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

v_0 = -3 m/s

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.

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brainly.com/question/5248528

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4 0
3 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

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 V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  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

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 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

8 0
3 years ago
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