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Diano4ka-milaya [45]
3 years ago
9

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

Physics
1 answer:
ruslelena [56]3 years ago
8 0

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

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lys-0071 [83]

Answer:

The object will have an upward acceleration

Explanation:

Let's consider the forces applied on the box. We have only two forces:

- The applied force of push, F_p, downward

- The force of gravity, W, (also known as weight of the object), downward

Therefore, the net force on the box is:

F_{net}=F_p -W

Here, we know that force applied is equal or greater than the weight, so

F_p \geq W

And therefore the net force is greater than zero:

F_{net}\geq 0 (1)

According to Newton's second law of motion, the net force on the box is equal to the product between its mass and its acceleration:

F_{net}=ma

where

m is the mass of the box

a is its acceleration

Given (1), this means that

a\geq 0

Therefore, the box will have an upward acceleration.

In this case force example we have:

F_p = 100 N\\W = 40 N

So the mass of the box is

m=\frac{W}{g}=\frac{40}{10}=4 kg

So the net force is

F_{net}=F_p-W=100-40=60 N

And the acceleration is

a=\frac{F_{net}}{m}=\frac{60}{4}=15 m/s^2

4 0
3 years ago
Which of the preceding statements is/are true?A) Helioseismology is the study of the differential rotation and magnetic field of
Nutka1998 [239]

Answer:

A and C

Explanation:

Helioseismology is a branch of science that makes use of  differential rotation and magnetic field to study he structure of the sun.

A filtergram represents the photo of the surface of the sun that is made using high energy particles being accelerated by the sun's magnetic field. <em>Hence, it can only be used to study the surface and not the layers below the photosphere.</em>

The chromosphere of the sun has a temperature between 6,000 °C to about 20,000°C while the photosphere has a temperature between 4,230 and 5,730 °C.<em> Hence, the chromosphere has a higher temperature.</em>

The only true options are A and C.

8 0
3 years ago
The effect of a particle in a fluid attaining its terminal velocity is that the?​
eduard

Answer:

In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving. ... At this point the object ceases to accelerate and continues falling at a constant speed called the terminal velocity (also called settling velocity).

5 0
4 years ago
Ce unitate de masura are indicele de REFRACTIE?
ivolga24 [154]

Answer:se obesrva ca indicele de refractie NU are unitate de masura, este adimensional. Se exprima print-un raport, de exemplu 4/3, 1,5 etc.

Explanation:

3 0
2 years ago
Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would
Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight P_{1} = 30 W

Power of starter P_{2} = 2500 W

Voltage of headlight and starter V = 12 V

From equation of power,

 P = \frac{V^{2} }{R}

 R = \frac{V^{2} }{P}

For finding the resistance of headlight and starter,

⇒ For headlight,

 R_{1}  = \frac{144}{30} = 4.8 Ω

⇒ For starter,

R_{2} = \frac{144}{2500} = 0.057 Ω

Since equivalent resistance,

R_{eq} = R_{1} + R_{2} + ........

R_{eq} = 4.8 +0.057 = 4.857 Ω

So power in series is given by,

 P_{s } = \frac{V^{2} }{R_{eq} }  = \frac{144}{4.857}

 P_{s } = 29.64 W

8 0
3 years ago
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