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3 years ago

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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How much force does a soccer goalie

Tems11 [23]

Answer:

2 N

Explanation:

From the question, it's given that

Mass m = 0.2 kg

Acceleration a = 10 m/s^2

The force a soccer goalie experience when stopping a ball will be equal to the force at which the ball is being kicked. This is

F = ma

Substitute all the parameters into the formula

F = 0.2 × 10

F = 2 Newton.

3 0

3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

7 0

3 years ago

Read 2 more answers

5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur

Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

Л = 1 - $\frac{Tl}{Th}$

where Tl = 300 k

Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 ) :

= P3v3 = mRT3 ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

= (1076.25 * 0.4 ) / 1237.2

= 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

= 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below

6 0

3 years ago

A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

vazorg [7]

Answer:

Explanation:

It is 8z

3 0

3 years ago

One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri

maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0

3 years ago