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mixas84 [53]
3 years ago
7

Why were quasars and active galaxies not initially recognized as being “special” in some way?

Physics
1 answer:
djyliett [7]3 years ago
8 0

Answer:

In the past, astronomers look into the sky and the universe as a whole with an idea that it is a place where stars are born, transition through their life stages and ultimately die and this is because they couldnt differentiate between stars, quasars and active galaxies because with advanced equipment, they all look similar but as they technological ages arrived, they were able see that they are not the same.

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A .2kg Basketball is pitched with a velocity or 40 m/s and then bed and into the picture with a velocity of 60 m/s. What is the
Tamiku [17]

Answer:

40kgm

Explanation:

∆p = m(v - u)

= 2(60 - 40)

= 2 × 20

= 40kgm/s

4 0
3 years ago
Define the law of universal gravitation in your own words.
Firdavs [7]
Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
7 0
3 years ago
If you put a object in front of the mirror why does it appear the opposite way
UNO [17]
It is because that is how mirrors work, they reflect light, and since we see objects because we are seeing the light these objects reflect, what is reflected back by the mirror is what we see.
4 0
3 years ago
A roller coaster's velocity at the top of the hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocit
svet-max [94.6K]

your answer is 8 m / sec²

5 0
2 years ago
You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m
sergey [27]

Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

  • mass of vehicle, m=1350\ kg
  • radius of curvature, r=71\ m
  • coefficient of friction, \mu=0.23

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

where:

\mu= coefficient of friction

N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

v=12.65\ m.s^{-1} is the maximum velocity at which the vehicle can turn without skidding.

5 0
3 years ago
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