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tresset_1 [31]
3 years ago
15

The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m

odule for landing on the surface of Mars. Use conservation of mechanical energy to answer these questions.
(a) As the lander is descending, if the pilot decides to shut down the engine when the lander is at a height of 1.3 m, (this may not be a safe height to shut down the engine) and the velocity of the lander (relative to the surface of the planet) is 1.3 m/s what will be velocity of the lander at impact? Note: g on the surface of mars is about 0.4 times that on the surface of the Earth.

(b) In the case of the lunar module an impact velocity of 3.0 m/s or less was essential for a safe landing. Assuming this to be the case for the Mars lander as well, at what maximum height could the pilot shut down the engines to ensure a safe landing. Assume the velocity v0 at the time the engine is shut down is 1.2 m/s.
Physics
1 answer:
lions [1.4K]3 years ago
8 0

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

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A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What
Ratling [72]

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

<u>V₀y = 0 m/s</u>

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

<u>t = 2.47 s</u>

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

<u>V₀ₓ = 61.86 m/s</u>

<u></u>

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

4 0
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What is the most important factor affecting the movement of water in an ocean?
arlik [135]
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The color of light most readily absorbed by water is _________.
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A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma
Ierofanga [76]

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter a=1.6m/sec^2

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So F=70\times 1.6=112 N

So external force will be 112 N

6 0
3 years ago
I got part c right but idk why the other parts are wrong HELP!
dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

J=\int F\Delta t

where

F is the magnitude of the force exerted on the object

\Delta t is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

B=0.14-0 = 0.14s\\b=8-5=3s

and whose height is

h=900 N

Therefore, the area (and the impulse) is

J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

J=F_{avg} \Delta t

where

F_{avg} is the average force exerted during the whole time \Delta t

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

\Delta t = 0.14 s is the time interval

Solving for the average force, we find

\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

J=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
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