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sesenic [268]
3 years ago
9

Describe succinctly the relationship between how far a galaxy is from us (its distance), versus how fast it is moving.

Physics
1 answer:
Eva8 [605]3 years ago
7 0

Answer:

Distance is directly proportional to the velocity

Explanation:

In 1929, Edwin Hubble's wrote an article that talked about relationship between the distance and recession speed/velocity of galaxies which led to what is known as the Hubble Law. This law states that galaxies are moving away from the earth at velocities proportional to their distances.

Thus is written as;

v = H_o•d

Where;

v is velocity

d is distance

H_o is Hubble's constant rate of cosmic expansion.

He came to this conclusion by generating a graph known as Hubble's classic graph which was a graph of observed velocity vs distance for nearby galaxies.

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8. A car moving at 35 m/s has 675 joules of KE. What is the mass of the car?
andreyandreev [35.5K]
Kinetic energy is the energy possessed by a body in motion while potential energy is the energy of a body at rest.
Kinetic energy is given by E=1/2MV² where M is the mass of the body while V is the velocity of the body. 
To get mass we can use the formula M= 2 Ek/V² (Making M the subject)
hence mass  = (2 ×675)÷35²
                     = 1.102 kg 
6 0
3 years ago
Hello please help i’ll give brainliest
Masteriza [31]

Answer:

D, the lithosphere. (CRUST AND UPPER MANTLE)

Explanation:

A tectonic plate (also called lithospheric plate) is a massive, irregularly shaped slab of solid rock, generally composed of both continental and oceanic lithosphere. Plate size can vary greatly, from a few hundred to thousands of kilometers across; the Pacific and Antarctic Plates are among the largest. Plate thickness also varies greatly, ranging from less than 15 km for young oceanic lithosphere to about 200 km or more for ancient continental lithosphere (for example, the interior parts of North and South America).

Information found on:

<u>https://pubs.usgs.gov/gip/dynamic/tectonic.html#:~:text=A%20tectonic%20plate%20(also%20called,both%20continental%20and%20oceanic%20lithosphere.&text=Continental%20crust%20is%20composed%20of,such%20as%20quartz%20and%20feldspar.</u>

3 0
3 years ago
Read 2 more answers
How does the period affects the centripetal force?
Anna71 [15]

Answer:According to the Equation (2), centripetal force is proportional to the square of the speed for an object of given mass M rotating in a given radius R.

Explanation:The Period T. The time T required for one complete revolution is called the period. For. constant speed. v = 2π r T holds.

8 0
3 years ago
Read 2 more answers
A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil
user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0
3 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
3 years ago
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