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3 years ago

You overhear two students discussing the topic of Doppler shift.

Physics

1 answer:

9966 [12]3 years ago

5 0

Answer:

agree with student 2, disagree with student 1

Explanation:

If you want to know if the wavelength of light was shifted you have to know the original wavelengths

Since we know the absorption spectrum for elements like hydrogen, we can look for these absorption lines in the star's spectra and figure out what direction these lines are shifted and tell if the star is moving away or towards us

The color of the star refers to the temperature of the star's surface which is not related to the doppler shift of the star

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the mo

Dennis_Churaev [7]

Answer:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

7 0

2 years ago

A person has a mass of 60 kg. What is the person’s weight in Newtons and in pounds?

liubo4ka [24]

Answer:

137.2 in pounds and in Newton's it's 588.399

3 0

3 years ago

A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o

Elden [556K]

The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where we have (choosing downard as positive direction)

s = 125 m is the vertical displacement of the cat

u = 0 is the initial vertical velocity

t is the time taken to reach the ground

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(125)}{9.8}}=5.05 s$

Now we can analzye the horizontal motion. Since the motion is uniform and the horizontal velocity is constant, it is given by

$v_x =\frac{d}{t}$

where

d = 286 m is the horizontal distance travelled

t = 5.05 s is the time taken

Solving the equation,

$v_x = \frac{286}{5.05}=56.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

masya89 [10]

Answer: the magnetic wave will travel out of the screen.

Explanation:

Electric field direction is perpendicular to the magnetic field direction. Both are also perpendicular to the direction of the particles.

Using right hand rule to solve this problem,

This pointed finger depicts the electric field direction which the curly fingers depict the direction of the magnetic field. The pointed thumb will depict the direction in which the wave travel. Which is out of the screen.

8 0

3 years ago