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3 years ago

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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Is this true of false Apples and Onions taste the same if you plug your nose?

xxMikexx [17]

I think yes because you won’t be able to smell

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3 years ago

One beam of electrons moves at right angles to a magnetic field. the force on these electrons is 4.9 x 10-14 newtons. a second b

diamong [38]

Below are the choices that can be found elsewhere:

A. (4.9 × 10-14 newtons) · tan(30°)
B. (4.9 × 10-14 newtons) · sin(30°) 
C. (4.9 × 10-14 newtons) · cos(30°) 
D. (4.9 × 10-14 newtons) · arctan(30°) 
E. (4.9 × 10-14 newtons) · arccos(30°)

Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)

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3 years ago

Where is Chemical Energy in a material stored ?

Sphinxa [80]

It is stored in the bonds between atoms

6 0

3 years ago

Water flows through a 2.5cm diameter pipe at a rate of

My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² = (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

= 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of a constriction in the pipe at the new pressure = 0.0174m = 1.74cm

6 0

2 years ago

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance

Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

t = 5 s is the time

So, the final velocity is

$v=0+(2)(5)=10 m/s$

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

v = 10 m/s is the final velocity

Solving for s,

$s=\frac{10^2-0^2}{2(2)}=25 m$

3 0

3 years ago