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Anettt [7]
3 years ago
14

A man with 60 years to live wants to visit a distant Galaxy which is 160,000 light years away. What must be his constant speed?

Physics
1 answer:
Assoli18 [71]3 years ago
5 0

The man's constant speed is 2.99 x 10⁸ m/s.

<h3>What is speed of light?</h3>

The speed with which the light ray travels in a medium with negative pressure or in vacuum.

This speed is the greatest speed in the universe. No matter can have speed greater than the speed of light in vacuum.

Given is a man with 60 years to live wants to visit a distant Galaxy which is 160,000 light years away.

If a man travels with the speed of light, he needs 160000 years to reach the galaxy, but visits in 60 years.

According to the relativistic theory,

160000 = 60 /√[1 - (v² /c²)]

where v is constant speed need to find and c is the speed of light in vacuum = 3 x 10⁸ m/s

Substitute into the formula, we get

160000 =  60 /√[1 - (v² /(3 x 10⁸)²) ]

Simplifying further

v² /c² = (1 - 1.406 x 10⁻⁷ )

Then,

v/c = 0.999

v = 0.999 x 3 x 10⁸ m/s

v = 2.99 x 10⁸ m/s

Thus, the constant speed of the  man is 2.99 x 10⁸ m/s.

Learn more about speed of light.

brainly.com/question/394103

#SPJ1

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A)   F_g = 4.05 10⁻⁴⁷ N, B)   F_e = 9.2 10⁻⁸N, C)    \frac{F_e}{F_g} = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

            F = G \frac{m_1m_2}{r^2}

let's calculate

            F = 6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }

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B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

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            F = 9 \ 10^9 \  \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}

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C) It is asked to find the relationship between these forces

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4 0
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A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f
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Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

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Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

Acceleration due to gravity is, g=9.8\ m/s^2

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

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From the free body diagram,

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