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matrenka [14]
3 years ago
5

A specific example of how light year is used in space exploration. Pick one object in space (the nearest star, the nearest galax

y, the edge of the universe, etc.) and tell how far this object is from earth and how long it would take light to travel there.
Physics
1 answer:
VladimirAG [237]3 years ago
5 0
Alpha Centauri A is about 93 million miles away from the Earth and it takes light 4.367 years to reach there.
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Pam rubs a balloon on her head to generate a static charge. She holds the balloon up against a wall. Which of the following desc
cestrela7 [59]

Answer:

c

Explanation:

6 0
3 years ago
Using our understanding of the law of gravity. What happens to the gravity as we triple the distance between two objects?
MA_775_DIABLO [31]

Answer:

1/9

Explanation:

<em>Newton’s Law of Universal Gravitation </em>

Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.

F = GMm/r²

where  

F - the gravitatioal force in Newtons,  

M   and m  -two masses in kilograms  

r  - the separation in meters.  

G  - the gravitational constant (6.674*10 ⁻¹¹ N (m/kg) ² )

Because of the magnitude of  G , gravitational force is very small unless large masses are involved.

So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance

F1 ∝ 1/r² ---------------(1)

F2 ∝ 1(3r)²

F2 ∝ 1/9r²--------------(2)

(2)/(1)

\frac{F_2}{F_1} =\frac{1}{9}\\ F_2 =\frac{F_1}{9}

From their you get as the distance tripled, Force reduce by a factor of 9(3³)

for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)

3 0
3 years ago
Boyle's Law mainly involves _______.
goblinko [34]
Your answer is B, gases
6 0
3 years ago
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m
denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0
2 years ago
The force required to stretch a Hooke’s-law
Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K


Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J


5 0
3 years ago
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