On Apollo Moon missions, the lunar module would blast off from the Moon's surface and dock with the command module in lunar orbit. After docking, the lunar module would be jettisoned and allowed to crash back onto the lunar surface. Seismometers placed on the Moon's surface by the astronauts would then pick up the resulting seismic waves.
Find the impact speed of the lunar module, given that it is jettisoned from an orbit 110 km above the lunar surface moving with a speed of 1600 m/s .
My Approach:
Ei = Ef
1/2*m*vi2 - (G*m*ME)/(radius of moon + orbital distance) = 1/2*m*vf2 ​- (G*m*ME)/r
=> (0.5 * m * 16002) - (6.67 * 10-11 * 7.35*1022 * m/(1737.4*103 +180*103) = (0.5*m*v^2) - (6.67*10-11 * 7.35*1022 * m/(1737.4*103 )

Answers

Answer 1

Answer:

Following are the solution to the given question:

Explanation:

For crashing speed, we can use energy conservation:

kinetic energy [tex]= \frac{1}{2}\times m \times v^2[/tex]  

potential energy [tex]= -\frac{GMm}{r}[/tex]

moon mass[tex]= 7.36\times 10^{22} \ kg[/tex]

Radius[tex]= 1738\ km[/tex]  

[tex]\to (K + U) \ orbit = (K + U)\ crash\\\\\to \frac{1}{2}\times m \times v_o^2 - \frac{GMm}{(1738000 + 110000)} \\\\ \to \frac{1}{2}\times m \times vc^2 - \frac{GMm}{1738000}[/tex]

Calculating the mass drop for the leave:

[tex]\to \frac{vo^2}{2} - \frac{GM}{1848000}\\\\ \to \frac{vc^2}{2} - \frac{GM}{1738000}[/tex]

Solve the value for

[tex]vc = \sqrt{(vo^2 +2\times GM \times(\frac{1}{1738000} - \frac{1}{1848000}))}\\\\vc = \sqrt{(1600^2 +2\times 6.67\times 10^{-11} \times 7.36 \times 10^{22}\times (\frac{1}{1738000} - \frac{1}{1848000}))}\\\\vc = 1701 \ \frac{m}{s}\\\\[/tex]  

The approach is correct but misrepresented in replacing 180 km instead of 110 km.


Related Questions

mdjxjxjcjfkfjjdksklqlakzjxjxkkskakMmznxkxkdkd?​

Answers

Answer:

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

I hope this helped!+*

The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light.

a. True
b. False

Answers

Answer:

False

Explanation:

No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.

9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?

Answers

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

A winch is capable of hauling a ton of bricks vertically two stories (6.35 m ) in 24.5 s .
If the winch’s motor is rated at 5.80 hp , determine its efficiency during raising the load.

Answers

Answer: 84 %

Explanation:

1.) Calculate the mass of a solid gold rectangular bar that has dimensions lwh = 4.30 cm ✕ 14.0 cm ✕ 27.0 cm. (The density of gold is 19.3 ✕ 103 kg/m3.)
kg


2.)A brass ring of diameter 10.00 cm at 17.3°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 17.3°C. Assume the average coefficients of linear expansion are constant.

(a) To what temperature must the combination be cooled to separate the two metals?


(b) What if the aluminum rod were 10.06 cm in diameter?

Answers

Answer:

1) m = 0.3137 kg

2a)T_f = -181.7°C

2b) T_f = -1176.97°C

Explanation:

1) We are given;

Length; l = 4.30 cm = 0.043 m

Width; w = 14.0 cm = 0.014 m

height; h = 27.0 cm = 0.027 m

density of gold; ρ = 19.3 × 10³ kg/m³

Formula for the density is known as;

ρ = mass/volume

Thus;

m =ρV

m = 19.3 × 10³ × (lwh)

m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)

m = 0.3137 kg

2a) We are given;

Diameter of brass; L_br = 10 cm

Diameter of aluminum; L_al = 10.01 cm

Now, to some for change in temperature we will use the formula;

L_f = L_i + αL_i(Δt)

Where α is coefficient of expansion.

Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.

Thus;

L_f(brass) = L_f(aluminium)

From table attached, α_brass ≈ 19 × 10^(-6) /°C

Also, α_aluminium ≈ 24 × 10^(-6) /°C

Thus;

L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))

Similarly,

L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Since L_f(brass) = L_f(aluminium), then;

10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Rearranging, we have;

10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.01 = Δt(-50.24 × 10^(-6))

Δt = 0.01/(-50.24 × 10^(-6))

Δt ≈ -199°C

Thus, temperature at which the combination must be cooled to separate the two metals is;

T_f = T_i + Δt

T_f = 17.3 + (-199)

T_f = -181.7°C

2b) Diameter of aluminum is now;

L_al = 10.06 cm

Thus;

10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.06 = Δt(-50.24 × 10^(-6))

Δt = 0.06/(-50.24 × 10^(-6))

Δt = -1194.27°C

T_f = 17.3 + (-1194.27)

T_f = -1176.97°C

Describing a Wave
What does a wave carry?

Answers

Answer:

Waves carry energy from one place to another.

Explanation:

Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.

It carries energy from one area to another

A candle is placed 50 cm from a diverging lens with a focal length of 28. What is the image distance in cm.

Answers

Answer:

v = -17.94 cm

Explanation:

Given that,

The candle is placed at a distance of 50 cm, u = -50 cm

The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)

We need to find the image distance. We know that the lens formula is as follows:

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]

So, the image distance is equal to 17.94 cm.

The law of conservation of angular momentum states that if no external force acts on an object, then its angular momentum does not change. true or false

Answers

Answer:

the answer is false.

Explanation:

i took the test and it is false trust me!!!!!!!!!

Select the correct answer.
Each square dance begins with what?
A. Handshake
B. Dosado
C. Bow or curtsy
D. Promenade

Answers

Answer:

C

Explanation:

The men bow to the women and the women curtsy to the men.

(took on test and got it right)

10 POINTS! SPACE QUESTION!!

Answers

Answer; they are larger and made of rocky material

what does loudness of a sound depend on?​

Answers

Answer:

Amplitude

Explanation:

The loudness of a sound depends on the amplitude of vibration producing the sound

the answer amplitude. The amplitude of a periodic variable is a measure of its change in a single period. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values. In older texts, the phase of a period function is sometimes called the amplitude.

What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?

Answers

im confused hold on imma send you a link to the answerExplanation:

Momentum
Project: Egg Drop
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Answers

Answer:

get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it

Explanation:

Materials For Project:Four sheets of 8 1/2 x 11 inch paper + One meter of masking tape + One resealable plastic bag (sized for an egg) + One meter of string + Three Straws + Eggs + White school glue. Draw Three Sketches: of your ideas for the egg do drop and NOT crack. Then, claim the advantages and disadvantages for each sketch. (If you don't know how to draw, try your best or notify a teacher.) FINAL SKETCH: After drawling the three sketching pick the best one for your final work. Explain why you chose that sketch.

-Email your teacher if you are still confused.

Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tuning fork and tune the piano to match, but her G4 tuning fork is missing! Instead, she plays the errant key along with her F4 tuning fork (which has a frequency of 349.2 Hz), displays the resulting waveform on a handheld oscilloscope, and measures a beat frequency of 76.7 Hz. Then, she plays the errant key along with her A4 tuning fork (which has a frequency of 440.0 Hz) and measures a beat frequency of 14.1 Hz.
What frequency is being played by the out-of-tune key ?
a. 363.3 Hz
b. 451.1 Hz
c. 33.9 Hz
d. 272.5 Hz
e. 425.9 Hz

Answers

Answer:

e. 425.9 Hz

Explanation:

The computation of the frequency is being played by the out-of-tune key is shown below;

Given that

Δf1  = x - 349.2 = 76.7.........(1)

Δf2 = 440 - x  = 14.1......(2)

Now solve (1) and (2)

(440 - x) - x + 349.2 = 14.1 - 76.7

789.2 + (-2x) = -62.6

x = 425.9 Hz

Hence, the frequency is being played by the out-of-tune key is 425.9 Hz

Therefore the option e is correct

Which investigation BEST measures the gravitational force on a
toy car?
A. rolling the car down a steep ramp and measuring time
B. using a spring scale and measuring the weight of the car
C. pushing the car and measuring how far it travels before it stops
D. throwing the car in the air and measuring how far it goes before coming
down

Answers

Answer:

B : using a spring scale and measuring the weight of the car

Explanation:

1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20

Answers

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

Which one of the following statements concerning the magnetic field inside (far from the surface) a long, current-carrying solenoid is true?
1) The magnetic field is zero.
2) The magnetic field is independent of the number of windings.
3) The magnetic field varies as 1/r as measured from the solenoid axis.
4) The magnetic field is independent of the current in the solenoid.
5) The magnetic field is non-zero and nearly uniform.

Answers

Correct answe: 5 (the magnetic field is non-zero and nearly uniform)

Which of the following quantities are unknown? initial separation of the particles final separation of the particles initial speed of the proton initial speed of the alpha particle final speed of the proton final speed of the alpha particle mass of the proton mass of the alpha particle charge of the proton charge of the alpha particle Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if A, C, and D are unknowns, enter ACD.

Answers

ALL of the quantities are unknown, because you haven't bothered to tell us anything that's known.

HURRY IM TIMED

How can you make people feel inspired?

By leading them on an emotional journey through various states to inspiration
By talking about something that interests you
By proving yourself to be a trustworthy speaker
By making them laugh and feel comfortable

Answers

Answer:

By talking about something that interesto you’

sorry if wrong

Explanation:

A toy car rolls down a ramp. Which force causes the car to move

Answers

Answer:

Gravity

Explanation:

Gravity pulls things down to earth and it is a force

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a period of 0.54 s . His older sister pulls the spring a bit more than intended. She pulls the animal 32 cm below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position. Part A If the animal does not hit anything on the way up, how far above its equilibrium position will it go

Answers

Answer:

the wooden animal will go 0.7068 m above its equilibrium

Explanation:

Given the data in the question;

mass of wooden animal m = 120 g = 0.12 kg

the animal oscillates up and down, T = 0.54 s

older sister pulls the animal 32 cm below its equilibrium position;

x = 32 cm = 0.32 m

g = 9.81 m/s²

We know that

k = mω²

where ω = 2π/T

So, k = m( 2π/T )²

we substitute

k = 0.12( 2π / 0.54 )²

k = 0.12 × (11.6355)²

k = 0.12 × 135.38486

k = 16.25 N/c

so Also,

kx²/2 = mgh

we solve for h

h = kx² / 2mg

we substitute

h = ( 16.25 × (0.32)²) / ( 2 × 0.12 × 9.81 )

h = 1.664 / 2.3544

h = 0.7068 m

Therefore, the wooden animal will go 0.7068 m above its equilibrium

A coil of resistance 100ohm is placed in a magnetic field of 1mWb the coil has 100 turns and a galvanometer of 400ohm resistance is connected in series with it. find the average emf and the current if the coil is moved in one tenth of a second from the given field to a field of 2.0mWb​

Answers

Answer:

C

Explanation:

C

Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?

Answers

Answer:

[tex]0.0035\ \text{m}[/tex]

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = 700 nm

m = Order = 1

We have the relation

[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]

The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].

What is the chemical formula for the molecule modeled?

Answers

Answer:

What is the chemical fórmula For the molecule modeled?

Explanation:

C6H12O2

Consider a wheel (solid disk) of radius 1.12 m, mass 10 kg and moment of inertia 1 2 M R2 . The wheel rolls without slipping in a straight line in an uphill direction 37◦ above the horizontal. The wheel starts at angular speed 12.0536 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops?

Answers

Answer:

d= 23.25 m

Explanation:

Assuming no other external forces acting on the disk, total mechanical energy must be conserved.Taking the initial height of the disk as the zero reference for the gravitational potential energy, initially. all the energy is kinetic.This kinetic energy is part translational kinetic energy, and part rotational kinetic energy, as follows:

       [tex]E_{o} = K_{transo} + K_{roto} (1)[/tex]

When the disk rolling uphill finally comes to an stop, its energy is completely gravitational potential energy, as follows:

       [tex]E_{f} = m*g*h (2)[/tex]

Since the angle with the horizontal of the track on which the disk is rolling, is 37º, we can express the height h in terms of the distance traveled d and the angle of 37º, as follows:

       [tex]h = d* sin 37 (3)[/tex]

Replacing (3) in (2):

       [tex]E_{f} = m*g* d * sin 37 (4)[/tex]

Since the wheel rolls without sleeping, this means that at any time there is a fixed relationship in the translational speed and the angular speed, as follows:

       [tex]v = \omega * R (5)[/tex]

For a solid disk, as mentioned in the question, the moment of inertia is just 1/2*M*R².The rotational kinetic energy of a rotating rigid body can be written as follows:

       [tex]K_{rot} = \frac{1}{2}* I * \omega^{2} (6)[/tex]

Replacing I from (6) and ω from (5), and remembering the definition of the translational kinetic energy, we can solve (1) in terms of v, m and r as follows:

       [tex]E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)[/tex]

Since (4) and (7) must be equal each other, we can solve for d as follows:

       [tex]d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)[/tex]

Replacing by the values, we finally get:

       [tex]d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.[/tex]

A 35.0 g bullet strikes a 50 kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and the bullet fly off together at 8.6 m/s. What was the speed of the ballot before it struck the lumbar? Define the bullet and the wood as a system

Answers

Answer:

12294.31 m/s

Explanation:

Momentum = (mass)(velocity)

Momentum before = Momentum after

(momentum of bullet)+(momentum of block)=(momentum of bullet and block)

0.035v+50(0)=(0.035+50)(8.6)

0.035v=430.301

v=12294.3142857m/s

Philosophy: The Big Picture Unit 8

How does pragmatism differ from the utilitarianism of the previous era?

A. Utilitarianism did not consider rights and pragmatism did.
B. Pragmatism is concerned with function, utilitarianism with happiness.
C. Utilitarianism was created in England and pragmatism came from Germany.
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.

Answers


D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. The time that elapses from when the ball passes the bottom of the window on the way up, to when it passes the top of the window on the way back down is 1.3 s.
What was the ball’s initial speed, in meters per second?

Answers

Answer:

[tex]u=14.48m/s[/tex]

Explanation:

From the question we are told that:

Height of window [tex]h=2m[/tex]

Height of window off the ground [tex]h_g=7.5m[/tex]

Time to fall and drop [tex]t=1.3s[/tex]

 

Generally the Newton's equation motion  is mathematically given by

 [tex]s=ut+\frac{1}{2}at^2[/tex]

Where

[tex]h=ut+\frac{1}{2}at^2[/tex]

[tex]2=u1.3-\frac{1}{2}*9.8*1.3^2[/tex]

[tex]2=u1.3-8.281[/tex]

[tex]u=7.91m/s^2[/tex]  

Generally the Newton's equation motion  is mathematically given by

[tex]2as=v^2-u^2[/tex]

Where

[tex]-2gh_g=v^2-u^2[/tex]

[tex]-2*9.8*7.5=(7.91)^2-u^2[/tex]

[tex]-147=62.5681-u^2[/tex]

[tex]u=\sqrt{209.5681}[/tex]

[tex]u=14.48m/s[/tex]

Therefore the  ball’s initial speed

[tex]u=14.48m/s[/tex]

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.

Answers

Answer:

[tex]0.0018833\ \text{m/s}[/tex]

Explanation:

[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]

[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

We have the relation

[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]

[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]

The required answer is [tex]0.0018833\ \text{m/s}[/tex].

what happens when water at 4° celsius is heated further?​

Answers

Answer:

please give me brainlist and follow

Explanation:

4 degrees C turns out to be the temperature at which liquid water has the highest density. If you heat it or cool it, it will expand. ... Ice floats on top of lakes, preventing evaporation (and convection in the frozen layer), and lakes stay liquid underneath, allowing fish and other life to survive.

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