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Angelina_Jolie [31]
2 years ago
7

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

s?
A.Wavelengths measured would match the actual wavelengths emitted.
B.Wavelengths measured would be shorter than the actual wavelengths emitted.
C.Wavelengths measured would be longer than the actual wavelengths emitted.

If a star is moving toward us in space, what would we observe about the wavelength from these stars?

A.Wavelengths measured would be shorter than the actual wavelengths emitted.

B.Wavelengths measured would match the actual wavelengths emitted.

C.Wavelengths measured would be longer than the actual wavelengths emitted.
Physics
1 answer:
dangina [55]2 years ago
6 0
1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.

2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
f'= \frac{c}{c+v_s} f_0
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.
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What are the components of friction?
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<h2>Answer:</h2>

<u>Friction:</u>

When an object slips on a surface, an opposing force acts between the tangent planes which acts in the opposite direction of motion. This opposing force is called Friction. Or in other words, Friction is the opposing force that opposes the motion between two surfaces.

The main component of friction are:

<u>Normal Reaction (R): </u>

Suppose a block is placed on a table in the above picture, which is in resting state, then two forces are acting on it at that time.

The first is due to its weight mg which is working from its center of gravity towards the vertical bottom.

The second one is superimposed vertically upwards by the table on the block, called the reaction force (P). This force passes through the center of gravity of the block.

Due to P = mg, the box is in equilibrium position on the table.

<u>Coefficient of friction ( </u>μ )<u>: </u>

The ratio of the force of friction and the reaction force is called the coefficient of friction.

Coefficient of friction, µ = force of friction / reaction force

μ = F / R

The coefficient of friction is volume less and dimensionless.

Its value is between 0 to 1.

<u>Advantage and disadvantage from friction force: </u>

  • The advantage of the force of friction is that due to friction, we can walk on the earth without slipping.
  • Brakes in all vehicles are due to the force of friction.
  • We can write on the board only because of the force of friction.
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4 0
3 years ago
The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a
zlopas [31]

Answer:

s = 9.6\times 10^{-12}kg/s

Explanation:

Given:

Solute Diffusion rate  = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel  =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel  =0.10 cm

let the Solute Diffusion rate  of new channel = s

now equating the diffusion rate per unit volume for both the channels

\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}

thus,

s = 9.6\times 10^{-12}kg/s

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3 years ago
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

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u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

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v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

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W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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