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vichka [17]
4 years ago
14

A parsec is an astronomical unit of distance where 1 parsec = 3.26 light years (1 light year equals the distance traveled by lig

ht in one year). If the speed of light is 186,000 mi/s, calculate the distance in meters of an object that travels 9.5 parsecs.
Physics
1 answer:
motikmotik4 years ago
5 0

Answer:

2932016616×10⁸ m

Explanation:

1 parsec = 3.26 light years

Speed of light = 3×10⁸ m/s

Distance = Speed × Time

Distance traveled by light in one year

3×10⁸×365.25×24×60×60

3.26 light years = 3.26×3×10⁸×365.25×24×60×60 = 308633328 ×10⁸m

⇒ 1 parsec = 308633328 ×10⁸m

9.5 parsecs = 9.5×308633328×10⁸

⇒9.5 parsecs = 2932016616×10⁸ m

Distance in meters of an object that travels 9.5 parsecs is 2932016616×10⁸ m

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Answer:

15.3

+1.285

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Answer = 16.585

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3 years ago
A planet outside of our solar system
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Answer:

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3 years ago
I need help with question 8 and 9
Wewaii [24]

8) The height of the shelf is 2.04 m

9) The gravitational potential energy is 100 J

Explanation:

8)

The gravitational potential energy of an object is given by:

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object, relative to the ground

For the object in this problem, we have

m = 10 kg

g=9.8 m/s^2

And we know that

PE = 200 J

Therefore we can re-arrange the formula to solve for h, the heigth of the object:

h=\frac{PE}{mg}=\frac{200}{(10)(9.8)}=2.04 m

9)

As before, the gravitational potential energy of the object is given by:

PE=mgh

where in this  case, we have:

m = 10 kg is the mass of the object

h = 2.04 m is the height of the object from the ground

While in this case, the acceleration of gravity on this planet is

g=4.9 m/s^2

Substituting,

PE=(10)(4.9)(2.04)=100 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0
4 years ago
A ray at which angle would produce the most glare? A. 59 degrees C. 25 degrees D. 37 degrees
Mkey [24]
<span>A. 59 degrees......
Hope it works</span>
5 0
3 years ago
Read 2 more answers
A small truck has a mass of 2085 kg. How much work is required to decrease the speed of the vehicle from 22.0 m/s to 13.0 m/s on
Alisiya [41]

Answer:

Work required is 328387.5 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 2085kg

Initial velocity, Vi = 13m/s

Final velocity, Vf =22m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

Where;

W is the work required.

Kf is the final kinetic energy possessed by the object.

Ki is the initial kinetic energy possessed by the object.

But Kinetic energy = ½MV²

W = ½MVf² - ½MVi²

Substituting into the equation, we have;

W = ½(2085)*22² - ½(2085)*13²

Simplifying the equation, we have;

W = 1042.5 * 484 - 1042.5 * 169

W = 504570 - 176182.5

W = 328387.5J

Therefore, the work required to decrease the speed of the vehicle is 328387.5 Joules.

8 0
3 years ago
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