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Alex777 [14]
3 years ago
12

The two populations of stars in our galaxy are actually quite similar in many of their characteristics. True False

Physics
1 answer:
stich3 [128]3 years ago
8 0

Answer:

False

Explanation:

The two population of stars in our galaxy are nowhere near similar in many characteristics.

For instance, the stars in the milky way galaxy are about 2/3 of all observable stars. Yes, the populations of stars in the milky way and outside of it are both large, but the population in the milky way galaxy is double that of elsewhere.

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What is the position and kind of image produced by the lens below?
Yanka [14]
The answer is convex image
3 0
2 years ago
A 5.5kg radio is pushed across the table. If the acceleration is 5m/s to the right, find the net force exerted on the radio.
Sholpan [36]

Answer:

Net force exerted on the radio is 27.5 Newton.

Given:

Mass = 5.5 kg

Acceleration = 5 \frac{m}{s^{2} }

To find:

Force exerted on the radio = ?

Formula used:

F = ma

Where F = net force

m = mass

a = acceleration

Solution:

According to Newton's second law of motion,

F = ma

Where F = net force

m = mass

a = acceleration

F = 5.5 × 5

F = 27.5 Newton

Hence, Net force exerted on the radio is 27.5 Newton.


4 0
3 years ago
4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
11111nata11111 [884]

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

3 0
3 years ago
A bubble of air is rising up through the ocean. When it is at a depth of 20.0 m below the surface, where the temperature is 5.00
kotegsom [21]

Answer:

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P(h)= Po + ρgh

where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)

replacing values

P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature

therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po)(V/Vo)=(T/To)

or

V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³

V = 0.253 cm³

3 0
3 years ago
please go to my profile and answer the work and energy questions, i only have 20 minutes left to finish it
Alexeev081 [22]

Answer:

Okay I'll do it right now

Explanation:

:)

7 0
3 years ago
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