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bekas [8.4K]
3 years ago
13

Galaxy A is located 200 million light-years from Earth, while galaxy B is located 600 million light-years from Earth. According

to the Hubble law, how does the motion of the two galaxies compare?
A. Galaxies A and B are moving away from Earth at the same rate.
B. Galaxy A is moving away from Earth three times as fast as galaxy B.
C. Galaxy B is moving away from Earth three times as fast as galaxy A.
Physics
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

C

Explanation:

I think c is correct due to reason given by you

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Which of the following statements is true regarding the use of refrigerants?
Damm [24]
B.

Refrigerants should be handled using approved equipment only.
5 0
3 years ago
Read 2 more answers
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
HELP HELP HELP ME!!!
zzz [600]
I believe the answer is C. Hope this helps!!

7 0
3 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec

(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

(c) We have given final angular velocity \omega _f=675rev/min

And \omega _i=79.9rev/min

Time t = 63 sec

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2

(d) Change in angle is given by

\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev

7 0
3 years ago
Which portion of the electromagnetic spectrum is used in a microscope?
erik [133]
Answer is C.
 
Optical microscope involves passing visible light transmitted through or reflected from the sample through a single or multiple lenses to allow a magnified view of the sample.

8 0
3 years ago
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