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jek_recluse [69]
3 years ago
15

How are light years used to measure distances in the universe

Physics
1 answer:
FinnZ [79.3K]3 years ago
4 0
A light year is a unit of distance. It is a distance that light can travel in a years time which is six trillion miles. It is used to measure the distances in space. To take one example, the distance to the next nearest big galaxy, the Andromeda Galaxy, from earth is 21,000,000,000,000,000,000 km. 
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You might be interested in
By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles
Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r²   -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

<u>F' = (4/9)F</u>

7 0
3 years ago
Can power be negative in physics? I am doing a lab in which I run down the stairs as quickly as possible. They are 5.4m tall and
RUDIKE [14]

Answer:

Yes.

Explanation:

A negative power would just represent a loss of power. So in your case it lost -1252.16 W

8 0
3 years ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
Hurryyyyy plzzzzzz..........the one in the middle...
zheka24 [161]

Answer:

Friction force is independent of the direction of the contacting surfaces

Explanation:

It can go any way depending on how much force is being out on it.

5 0
3 years ago
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
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