1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
o-na [289]
3 years ago
8

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

lar and 430 km above the surface of the moon, where the acceleration due to gravity is 1.08 m/s2. the radius of the moon is 1.70 â 106 m?
Physics
1 answer:
almond37 [142]3 years ago
6 0
Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude (h=430 km=4.3 \cdot 10^5 m) of the orbit:
r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, g=1.08 m/s^2. So we can write
g=a_c= \frac{v^2}{r}
where a_c is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s

part b) The orbit has a circumference of 2 \pi r, and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h
So, the period of the orbit is 2.45 hours.
You might be interested in
1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?
shepuryov [24]

1.

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N


2.

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg


3.

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N


7 0
3 years ago
Match each statement with the characteristic of scientists.
Dimas [21]

Answer:

Options B, A, D, C

Explanation:

When a scientists, let's say Roberto wonders if the presence of other elements also affects the color of a flame, he can decide to prove this through a study. Therefore, in chemistry class, Roberto sees that traces of lithium makes a flame appear bright red. Subsequently, Roberto designs an experiment to test flame color in the presence of different elements and finally Roberto's friend tells him the color of a flame cannot be changed, but Roberto is still unsure.

5 0
3 years ago
Read 2 more answers
A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

7 0
3 years ago
Please answer D in the image with an explanation
puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0
3 years ago
Three people pull simultaneously on a stubborn donkey. jack pulls eastward with a force of 92.5 n, jill pulls with 89.9 n in the
alekssr [168]
Jack------------ force of 92.5 n   eastward-------Fjack(X)=92.5 n   Fjack(Y)=0

<span>jill ------------------------------- force of 89.9 n   northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>

</span>jane -----------------------------force of 163 n   southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n

Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n

Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast

the magnitude of the net force the people exert on the donkey. is 294.80 n southeast
7 0
3 years ago
Other questions:
  • A graduate student has done a careful analysis of the spectrum of a star. While she has found lines from many elements, there wa
    12·1 answer
  • What is a <br> material that restricts the flow of electricity or thermal energy
    5·3 answers
  • A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire
    10·1 answer
  • Help with 2 Physics questions, WILL CHOOSE BRAINLIEST
    9·1 answer
  • What is the affect of applying an unbalanced force on an object?​
    10·1 answer
  • The mass of an object changes as the distance from the center of gravity changes. True False
    7·2 answers
  • A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the
    10·1 answer
  • What are 3 things that can affect gravity?
    10·1 answer
  • Consider these two characteristics.
    10·2 answers
  • 49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!