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Colt1911 [192]
3 years ago
10

A distant galaxy has a redshift z = 5.82 and a recessional velocity vr = 287,000 km/s (about 96% of the speed of light.) Notice

that the equation z=vrcz=vrc does not hold true for recessional velocities approaching the speed of light. What is the distance to the galaxy in light years?
Physics
1 answer:
vlabodo [156]3 years ago
7 0

Answer: 4100 Mpc

Explanation:

Since H o = 70 km/s/Mpc

Redshift z = 5.82

Recessional velocity vr = 287,000 km/s

Then, the distance to the galaxy in light years will be:

= Recessional velocity / H o

= 287000 / 70

= 4100 Mpc

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An astronaut is walking in space. Which of these would have the greatest speed as observed by the astronaut?
Aleks04 [339]
The answer is C) an electromagnetic wave

An electromagnetic wave, which includes electromagnetic radiation such as visible light, moves the fastest of all of the options listed by a significant margin, especially through space. In fact, light travelling through space is technically the theoretical limit of how fast something can travel. 
4 0
3 years ago
Read 2 more answers
What is air pressure?
dolphi86 [110]

Answer:

A. Earth's gravity pulling down on air molecules

Explanation:

Air pressure refers to the weight of the air per unit surface area. It is the amount of gravitational force which is pulling down the molecules of air.

The common unit of air pressure is: Pascal, atm

1 atm = 101325 Pa

As the column of the air above increases, the air pressure increase. This is because with the increase in amount of air, the weight increase of the air increases. This is the reason a diver feels immense pressure in the sea and cooking takes a lot of time on hilly areas because of low air pressure.

5 0
3 years ago
A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?
ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

6 0
3 years ago
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A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
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