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SVETLANKA909090 [29]
3 years ago
6

A starlike object seen on deep sky photographs coincides with an intense radio source and has a spectrum in which the characteri

stic lyman pattern of hydrogen spectral lines has been shifted from the ultraviolet to the visible spectral range. what is this object?
Physics
1 answer:
liberstina [14]3 years ago
6 0
A quasar - <span>Quasars are very distant objects. They  are really bright masses of energy and light. The word quasar is actually short for quasi-stellar radio source or quasi-stellar object. This high amount of energy in such a small area can only be formed by matter falling towards a Black Hole. That's why quasars are believed to be huge black holes  in the center of young galaxies.
</span><span />
You might be interested in
The volume of a fixed mass of gas is directly proportional to its
Leni [432]

Answer: Charles's law

Explanation:

Charles's law is one of the gas laws, and it explains the effect of temperature changes on the volume of a given mass of gas at a constant pressure. Usually, the volume of a gas decreases as the temperature decreases and increases as the temperature also increases.

Mathematically, Charles's law can be expressed as:

V ∝ T

V = kT or (V/T) = k

where v is volume, T is temperature in Kelvin, and a k is a constant.

7 0
3 years ago
A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0
3 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ball B has a mass of 3 kilograms. The initial v
ratelena [41]
1. The balls move to the opposite direction but the same speed. This represents Newton's third law of motion.
2. The total momentum before and after the collision stays constant or is conserved.
3. If the masses were the same, the velocities of both balls after the collision would exchange.
4 and 5. Use momentum balance to solve for the final velocities.
8 0
3 years ago
Read 2 more answers
Which is a form of precipitation?<br> A. Runoff<br> B. Rain<br> C. Gas<br> D. Evaporation
saw5 [17]

Answer:

D. Evaporation

Explanation:

Just did it made 100%

7 0
3 years ago
Read 2 more answers
A 1500 kg truck travelling north at 60 km/hr collides with a 1200 kg car moving east at 15km/hr. If the two cars remain locked t
Alenkinab [10]

Answer:

33.33j+6.67i km/hr

Explanation:

From the law of conservation of momentum,

Applying,

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.

Note: let j represent the north, and i  represent the east

From the question,

Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i

Substitute these values into equation 1

1500*60j+1200*15i = V(1500+1200)

90000j+18000i = 2700V

V = (90000j+18000i)/2700

V = 33.33j+6.67i km/hr

6 0
3 years ago
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