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1 year ago

There are planets in the universe without a parent star. most all stars have planets. most stars are binary. true false

Physics

1 answer:

Ghella [55]1 year ago

5 0

True : yes there are planets in the universe without a parent star called

rogue planets.

False : most stars are not binary

What is rogue planets ?

planets which do not orbits the planetary system is known as roque planets. For example, pluto. Rogue planets can be smaller and bigger than jupiter too.

Free-floating planets or rogue planets without orbits around stars may also harbor life, some scientists believe. Due to the gravitational effects of giant planets within their system, these planets get kicked out of their system like any other planets formed around stars.

no most stars are not binary, binary which means two stars orbit around the common center and approximately only 40 binary stars have been found till not in our solar system.

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A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

5 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

Tju [1.3M]

Answer:

675m

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 6m/s²

Time = 15s

Unknown:

Distance traveled by the body = ?

Solution:

To solve this problem; we use the expression;

S = ut + $\frac{1}{2}$ at²

Where u is the initial velocity

t is the time

a is the temperature

Insert the parameters and solve;

S = 0 x 15 + $\frac{1}{2}$ x 6 x 15²

S = 675m

5 0

2 years ago

An automobile is traveling away from Jill and towards Jack. The horn is honking, producing a sound wave.

wlad13 [49]

Answer:

a. The sound will travel at the speed to both Jill and Jack

b. Jack

Explanation:

a. Doppler effect describes how the frequency of a sound wave changes with regards to an observer that has relative motion to the sound source;

The Doppler effect is given by the following formula;

For a sound that is moving away, as observed by Jill, we have;

$f' = \dfrac{(v - v_0)}{(v + v_s)} \cdot f = \dfrac{v }{(v + v_s)} \cdot f$

For approaching sound, as observed by Jack, we have;

$f' = \dfrac{(v + v_0)}{(v - v_s)} \cdot f = \dfrac{v }{(v - v_s)} \cdot f$

Where;

f = The sound wave's actual frequency

f' = The frequency of the moving sound to the observer

v = The speed of the sound wave

v₀ = The observer's velocity = 0

$v_s$ = The velocity of the source (the automobile honking) of the sound wave

From the above equation, we have that the speed of sound, 'v', is the same to both the source, and the observer although the frequency, and therefore, the wavelength of the sound alternatively increases or decreases

b. From the Doppler effect equation, the person who will hear the highest frequency is given by the formula for the frequency when the sound is approaching the observer, which has the lower denominator

Therefore, given that the automobile is travelling towards Jack, jack will hear the higher frequency

5 0

3 years ago

What is the center of our solar system?

Yakvenalex [24]

The sun is the centre of our solar system

8 0

3 years ago

A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. What is the moment of inertia abou

frutty [35]

Answer:

I1 = 2/5 M1 R^2 for a sphere about its center

I2 = 2/5 M2 (2 R)^2 = 2/5 M2 R^ * 4 = 8/5 M2 R^2

Remember that M2 is greater than M1 by a factor 0f 2^3 = 8

Then I2 exceeds I1 by a factor of 32

6 0

2 years ago