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Galina-37 [17]
3 years ago
5

When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr

eely falls to the surface of the Moon.
i) How long does it take for the Eagle to touch down?
ii) What is the velocity of the lunar module when it hits the surface of the moon?
Physics
1 answer:
Scrat [10]3 years ago
7 0

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

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Explanation:

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All metals are malleable.

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Suppose the moon were held in its orbit not by gravity but by the tension in a massless cable. You are given that the period of
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Answer:

Tension in the cable = 200.655 × 10^(18) N

Explanation:

The tension in the cable will be the force on the string which is caused by when it holds against the centripetal acceleration.

Now, the formula for centripetal acceleration is given by;

a = v²/R

Now, let's find the velocity.

The circumference is 2πR

We are given that R = 3.85 x 10^(8) m

So, Circumference = 2 × π × 3.85 x 10^(8) = 24.19 × 10^(8) m

We are told the period of the moon orbit is 27.3 days

Distance travelled per day = Circumference/ period

Distance travelled per day = (24.19 × 10^(8))/27.3 = 88608058.608 m/day

Now, a day has 24 hours = 24 × 60 × 60 = 86400 seconds

Thus,

Distance travelled per seconds =

88608058.608/86400 = 1025.556 m/sec

So, from a_c = v²/r,

a_c = 1025.556²/3.85x10^8

a_c = 0.00273 m/s²

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3 years ago
If vector y = 21cm and vector Z=75 cm, what is vector x ?
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4 years ago
A certain sprinter has a top speed of 10.5 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able
Svet_ta [14]

Answer:

Total time = 10.667 seconds

Explanation:

For the first 12m

V² = u² + 2as

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V = u + at

10.5 = 0 + 4.59375t

t = 10.5/4.59375

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For the remaining race

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But a = 0

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Total time = 2.286 + 8.381

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