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3 years ago

What happens to light waves from a star as the star moves toward Earth?

Physics

2 answers:

Ksivusya [100]3 years ago

4 0

Answer:

The other person is right, the answer is they appear to shift towards blue.

Explanation:

Musya8 [376]3 years ago

3 0

This shifts the star’s spectral lines toward the blue end of the spectrum. If the star is moving away from us, it’s waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star’s spectral lines toward the red end of the spectrum.

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Question 15

Butoxors [25]

Jjdjdjfjififififkjfndndj

5 0

3 years ago

Determine the speed of the 50-kg cylinder after it has descended a distance of 2 m, starting from rest. Gear A has a mass of 10

Naddika [18.5K]

This question is incomplete, the missing image is uploaded along this answer below.

Answer:

the speed of the 50-kg cylinder after it has descended is 3.67 m/s

Explanation:

Given the data in the question and the image below;

relation between velocity of cylinder and velocity of the drum is;

V $_D$ = ω $_c$ × r $_c$ ----- let this be equ 1

where V $_D$ is velocity of cylinder, ω $_c$ is the angular velocity of drum C and r $_c$ is the radius of drum C

Now, Angular velocity of gear B is;

ω $_B$ = ω $_C$

ω $_B$ = V $_D$ / r $_c$ -------- let this equ 2

so;

V $_D$ / 0.1 m = 10V $_D$

Next, we determine the angular velocity of gear A;

from the diagram;

ω $_A$ ( 0.15 m ) = ω $_B$ ( 0.2 m )

from equation 2; ω $_B$ = V $_D$ / r $_c$

ω $_A$ ( 0.15 m ) = (V $_D$ / r $_c$ ) 0.2 m

substitutive in value of radius r $_c$ (0.1 m)

ω $_A$ ( 0.15 m ) = (V $_D$ / 0.1 m ) 0.2 m

ω $_A$ ( 0.15 ) = 0.2V $_D$ / 0.1

ω $_A$ = 2V $_D$ / 0.15

ω $_A$ = 13.333V $_D$ ----- let this be equation 3

To get the speed of the cylinder, we use energy conversation;

assuming that the final position is;

T₁ + ∑ $U_{1-2$ = T₂

0 + m $_D$ gh = $\frac{1}{2}$ m $_D$ V² $_D$ + $\frac{1}{2}I_A$ ω² $_A$ + $\frac{1}{2}I_B$ ω² $_B$

m $_D$ gh = $\frac{1}{2}$ m $_D$ V² $_D$ + $\frac{1}{2}$ (m $_A$ k $_A$ ²)(13.333V $_D$ )² + $\frac{1}{2}$ (m $_B$ k $_B$ ²)(10V $_D$ )²

we given that; m $_D$ = 50 kg, h = 2 m, m $_A$ = 10 kg, k $_A$ 125 mm = 0.125 m, m $_B$ = 30 kg, k $_B$ = 150 mm = 0.15 m.

we know that; g = 9.81 m/s²

so we substitute

50 × 9.81 × 2 = ( $\frac{1}{2}$ × 50 × V $_D$ ²) + $\frac{1}{2}$ ( 10 × (0.125)² )(13.333V $_D$ )² + $\frac{1}{2}$ ( 30 × (0.15)²)(10V $_D$ )²

981 = 25V $_D$ ² + 13.888V $_D$ ² + 33.75V $_D$ ²

981 = 72.638V $_D$ ²

V $_D$ ² = 981 / 72.638

V $_D$ ² = 13.5053

V $_D$ = √13.5053

V $_D$ = 3.674955 ≈ 3.67 m/s

Therefore, the speed of the 50-kg cylinder after it has descended is 3.67 m/s

7 0

3 years ago

A force does work on an object if a component of the force is

IrinaK [193]

Answer:

A force does work on an object if a component of the force is parallel to the displacement of the object.

Explanation:

Work, a measurement of energy is said to be done when a force applied to an object results in the movement of that object to a certain distance and direction. Force is the act of push or pulls occurs on an object as a result of the interaction between that object with another one and displacement is the distance and direction covered by that object as a result of the force applied on it.

The work done (W) by a constant force (F) is equal to the product of the force in the direction of displacement of the object and the distance (d) moved by the object i.e., W = F * d.

The angle between the displacement and the force is θ, then the work done, W = Fd cos θ ........ (1)

Positive work - Force acts in the same direction with respect to the displacement of the object. Here, θ is zero, so cos θ i.e., cos 0 is 1. Therefore, from the equation (1), W = Fd (i.e., work done by the force is positive).

Negative work - Force acts in the opposite direction with respect to the displacement of the object. Here, θ is 180°, so cos θ i.e., cos 180° is -1. Therefore, from the equation (1), W = -Fd (i.e., work done by the force is negative).

If a force is applied to an object and it does not move, then the work done is zero i.e., W = F * 0 = 0. Also, if the force and displacement are at right angle to each other, then θ is 90°. Therefore, from the equation (1), W = 0 since cos 90° is zero.

7 0

3 years ago

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

FrozenT [24]

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = $M_{0} V$

where $M_{0}$ is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = $M_{0}/ \sqrt{1 - \beta ^{2} }$

where

$\beta = V/c$

c is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> $M_{0}V/ \sqrt{1 - \beta ^{2} }$

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75 $M_{0} V$ = $M_{0}V/ \sqrt{1 - \beta ^{2} }$

the equation reduces to

1.75 = $1/ \sqrt{1 - \beta ^{2} }$

square both sides of the equation, we have

3.0625 = 1/ $(1 - \beta ^{2} )$

3.0625 - 3.0625 $\beta ^{2}$ = 1

2.0625 = 3.0625 $\beta ^{2}$

$\beta ^{2}$ = 0.67

β = 0.819

substitute for $\beta = V/c$

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

6 0

3 years ago

What is the difference between center of mass and center of gravity?

sweet [91]

Answer:

Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.

5 0

3 years ago

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