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Charra [1.4K]
3 years ago
7

The distance light travels in one year is called a light year. It is about 5.88 trillion miles and is used to measure stellar di

stances. At present, it is thought that the edge of the universe is about 14 billion light years from earth. Traveling at the speed of light, how long would it take you to reach the present edge of the universe? O 600 trillion miles O 14 billion miles O 14 billion years O 5.88 trillion light years
Physics
1 answer:
VikaD [51]3 years ago
3 0

Answer:

14 billion years

Explanation:

The distance of the edge of the universe = 14 billion light years

Light year is used to measure the distance which is equal to the distance traveled by the light at the speed of approx 3\times 10^8\ m/s in one year.

It means that to reach the edge of the universe which is 14 billion light years apart , if one is travelling at the speed of light which is 3\times 10^8\ m/s, it will take 14 billion years to reach there.

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Therefore, the relationship are as follows:
mass and gravity are inversely proportional 
mass and weight are directly proportional
weight and gravity are directly proportional 
7 0
3 years ago
Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift
sammy [17]

Answer:

The doppler effect equation is:

f' = \frac{v +v0}{v - vs}*f

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) -  (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0
3 years ago
Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are
Mars2501 [29]

Answer:

 θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

F = \sqrt{F_1^2 + F_2^2}

F = \sqrt{22.8^2 + 16.6^2}

F = \sqrt{795.4}

      F = 28.20 N

direction

\theta = tan^{-1}(\dfrac{F_2}{F_1})

\theta = tan^{-1}(\dfrac{16.6}{22.8})

\theta = tan^{-1}(0.728)

       θ = 36°

5 0
3 years ago
For a mass oscillating on a spring at what positions are (a) velocity and (b) acceleration of the mass have maximum valeus?
EleoNora [17]

Answer:

a)At the mean position

b)At the extremes positions

Explanation:

Given that mass is having oscillation motion.

We know that

1. At the mean position -The velocity of the mass is maximum and the acceleration of the mass is minimum.The net force on the mass will be zero.

2. At the extreme position-The velocity of the mass is minimum and the acceleration of the mass is maximum.The net force on the mass will not be zero.

Therefore

a)At the mean position

b)At the extremes positions

3 0
3 years ago
Helen adjusts the armature of an electric generator by increasing the number of coils around the iron core. What is Helen most l
Luba_88 [7]

Answer:

D

Explanation:

Got it right on the edge quiz

8 0
3 years ago
Read 2 more answers
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