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2 years ago

What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?

Physics

1 answer:

lora16 [44]2 years ago

3 0

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

$v = H_o D\\\\H_o = \frac{v}{D}$

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

$H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}$

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

2 years ago

The graph shows the federal budget from 1980 to 2010. In which period did the federal budget show the greatest deficit?

kodGreya [7K]

Answer:

2000 to 2010

Explanation:

sorry I'm just guessing

8 0

2 years ago

A ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st

coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0

2 years ago

16. An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

V125BC [204]

D. 128.25 because force=mass x acceleration

8 0

3 years ago

Read 2 more answers

How does friction affect the kinetic energy of an object?

Law Incorporation [45]

Answer: Friction also prevents an object from starting to move, such as a shoe placed on a ramp. When friction acts between two surfaces that are moving over each other, some kinetic energy is transformed into heat energy. Friction can sometimes be useful.

Explanation:

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3 years ago