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Galina-37 [17]
2 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]2 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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LW Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of wat
ruslelena [56]

Answer:

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Explanation:

given data

dia = 2 in = 0.166 feet

area = 26 km² = 26 × 247.105 acre = 6424.73 acre

to find out

volume of water

solution

we know volume of water is express as

volume = area × dia     ....................1

put here value in equation 1

volume = 6424.73 acre × 0.166 feet

volume = 1066.50

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4 0
3 years ago
A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= 10+1.50

= 11.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{11.5}

      =26.08 \ s

Throughout the opposite direction, when the boat seems to be travelling then,

= 10-1.50

= 8.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{8.5}

      =35.29 \ s

hence,

The time taken by the boat will be:

= 26.08+35.29

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8 0
2 years ago
A rock is thrown down with an initial speed of 25.3 m/s from a bridge to the water below. It takes 2.29 s for the rock to hit th
Rzqust [24]
<h2>Height of bridge is 83.66 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 25.3 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 2.29 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 25.3 x 2.29 + 0.5 x 9.81 x 2.29²

                      s = 83.66 m

      Height of bridge is 83.66 m

7 0
3 years ago
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