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4 years ago

The farther a star is away from earth, the more it is

Physics

2 answers:

Darya [45]4 years ago

8 0

The more it is Red shifted.

ella [17]4 years ago

4 0

Answer:

The farther a star is away from earth, the more it is red shifted

Explanation:

Our universe is expanding. The galaxies move away from each other. The farther the object is from earth, greater is the velocity with which it moves away.

When the light source moves away from us, their frequency appears to be lower than their real frequency. This is called Doppler effect.

Higher the velocity of the receding source, greater the reduction in frequency. Since the red light has lower energy in the visible light spectrum. The starts moving away tends to appear red rather than its original color.

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Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

$T_e=2\pi \sqrt{\frac{L}{g_e}}$

where

$g_e = 9.8 m/s^2$ is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

$T_m=2\pi \sqrt{\frac{L}{g_m}}$

where

$g_m = 1.6 m/s^2$ is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

$\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}$

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0

3 years ago

A 5.0 kg medicine ball and a 7.0 kg medicine ball are the same size and shape. If you were in outer space, how would you determi

Scrat [10]

Answer:
The objects can be distinguished by their weight instead of by mass.

Explanation
Because mass is constant, the two objects cannot be distinguished by mass.

However, gravitational acceleration varies in outer space. Therefore the heavier mass will register a higher reading on a weighing scale.

Note that an object of mass M weighs Mg, where g = acceleration due to gravity.

5 0

3 years ago

For a person with a Near Point (NP) = 45 cm, what would be his prescription's lens power = _________ diopters A) 1.08 diopters

natita [175]

Answer:

The power of lens is 2.22 D.

Explanation:

Given that,

Distance of image v= 45 cm

We know that,

Distance of object u= ∞

Power of lens is the reciprocal of focal length.

$P=\dfrac{1}{f}$

We need to calculate the power of lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{45}+\dfrac{1}{\infty}$

$\dfrac{1}{f}=\dfrac{1}{0.45}+0$

$P=\dfrac{1}{f}=2.22$

Hence, The power of lens is 2.22 D.

7 0

3 years ago

What is the term used for the rate at which distance is traveled?

jeka94

Velocity

Hope this helps

7 0

4 years ago

Read 2 more answers

A wire with a circular cross section and a resistance R is lengthened to 9.66 times its original length by pulling it through a

damaskus [11]

Answer:

The resistance of the wire after it is stretched is 93.31R.

Explanation:

Resistance is the property of the material to oppose the current flow through it. It is given by the relation :

R = (ρl)/A

Here ρ is resistivity, l is length of wire and A is the area of the wire.

Let l₀, and A₀ are the original length and original circular cross section area of the wire. while l₁ and A₁ are the new length and new circular cross section area of the wire.

Volume of the original wire, V₀ = A₀ x l₀

Volume of the new wire, V₁ = A₁ x l₁

According to the problem. volume remain same. So,

V₀ = V₁

A₀ x l₀ = A₁ x l₁

It is given that l₁ = 9.66 x l₀. Substitute this value in the above equation;

A₀ x l₀ = A₁ x 9.66 x l₀

A₁ = A₀/9.66

Resistance of the original wire, R = (ρl₀)/A₀

Resistance of the new wire, R₁ = (ρl₁)/A₁

Substitute the value of l₁ and A₁ in the above equation.

R₁ = (ρ x l₀ x 9.66)/(A₀/9.66) = 93.31 x (ρl₀)/A₀

But (ρl₀)/A₀ = R. hence,

R₁ = 93.31 R

8 0

3 years ago