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Anuta_ua [19.1K]

3 years ago

13

Which statement does NOT explain the concept of the Doppler Effect? A) As an observer approaches a stationary sound, the sound w

aves are compressed, making the pitch higher. B) As an observer moves away from a stationary sound, the sound waves are stretched making the pitch lower. C) As a sound source approaches a stationary object, the sound waves are compressed, making the pitch higher. D) As a sound source moves away from a stationary object, the sound waves

2 answers:

Maurinko [17]3 years ago

8 0

D, a<span>s a sound source moves away from a stationary object, the sound waves. It is the opposite of what happens. </span>

Alex787 [66]3 years ago

7 0

D give credit to monkey

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1 m/s^2

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The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 Km/s.

tino4ka555 [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) the mass of the star Alpha is $M_{\alpha}$ = $7.80*10^{29} kg$

b) the mass of the star Beta is $M_{\beta}$ = $2.34*10^{30}kg$

c) the radius of the orbit of the orange star $R_{0}$ = $1.9*10^{9}m$

d) the radius of the orbit of the black hole $R_{B} = 34*10^{8}m$

e) the orbital speed of the orange star $V_{0} = 4.4*10^{2}km/s$

f) the orbital speed of the black hole $V_{B} =77 km/s$

Explanation:

The generally formula for orbital speed is given as $v =\frac{2\pi R}{T}$

where v is the orbital speed

R is the radius of the star

T is the orbital period

From the question we are given that

alpha star has an orbital speed of $V_{\alpha} = 36km/s = 36000m/s$

beta star has an orbital speed of $V_{\beta} = 12km/s = 12000m/s$

the orbital period is $T = 137d = 137(86400)sec$ 1 day is equal 86400 seconds

Making R the subject of formula we have for the radius of the alpha star as

$R_{\alpha} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(36000m/s)}{2\pi}$

$=6.78*10^{10}m$

for the radius of the Beta star as

$R_{\beta} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(12000m/s)}{2\pi}$

$= 2.26*10^{10}m$

Looking at the value obtained for $R_{\alpha} and R_{\beta}$

Generally the moment about the center of the mass are equal then

$M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$

Thus $3M_{\alpha} = M_{\beta} -------(1)$

Generally the formula for the orbital period is given as

$T =\frac{2\pi(R_{\alpha+R_{\beta}})^{3/2 }}{\sqrt{G(M_{\alpha}+M_{\beta})} }$

Then

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Where G is the gravitational constant given as $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}$

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(6.78*10^{10}m +2.26*10^{10}m)^{2}}{(137d*86400s/d)^{2}(6.67*10^{-11}m^{3}kg^{-1}s^{-2})}$

$M_{\alpha} + M_{\beta} = 3.12*10^{30}kg -------(2)$

Thus solving (1) and (2) equations

Mass of alpha star is $M_{\alpha}=7.80*10^{29}kg$

and the Mass of Beta is $M_{\beta} =2.34*10^{30}kg$

Considering the equation

$M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}$

Making $R_{\alpha} + R_{\beta}$ the subject

$R_{\alpha}+ R_{\beta} =\sqrt[3]{\frac{(M_{\alpha+M_{\beta}})T^{2}G}{4\pi^{2}} } -------(3)$

and considering this equation $M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}$ from above

we have that $R_{\beta} =\frac{M_{\alpha}R_{\alpha}}{M_{\beta}}$

Considering Question C

Let the Orange star be denoted by (0) and

Let the black-hole be denoted by (B)

And we are told from the question that

Mass of orange star $M_{0} = 0.67M_{sun}$ and

Mass of black hole $M_{B} =3.8M_{sun}$

And mass of sun is $M_{sun} = 1.99*10^{30}kg$

Then $R_{B} = [\frac{0.67M_{sun}}{3.8M_{sun}} ]R_{0} =0.176R_{0}--------(4)$

We are also given that the period is $T =7.75 days = 7.75 (86400s)$

Considering equation 3

$R_{0} + 0.176R_{0} = \sqrt[3]{\frac{(0.67+3.8)(1.99*10^{30}kg)(7.75(86400s))^{2} 6.67*10^{-11}Nm^{2}/s^{2}}{4\pi ^{2}} }$

Thus for V616 Monocerotis, $R_{0} =1.9*10^{9}m$

Considering equation 4

The black-hole is

$R_{B}= 0.176R_{0}= 0.176*1.9*10^9 =34*10^8m$

From the formula for velocity of $V_{0} = \frac{2\pi R_{0}}{T} = 4.4*10^{9}km/s$

the velocity of $V_{B} = \frac{2\pi R_{B}}{T} =77km/s$

8 0

3 years ago

When a snowball turns into a puddle of

In-s [12.5K]

When a snowball turns into a puddle of water, we know that (the snow ball gains energy and changes from a solid to liquid).

This is correct due to the fact that particles of the snowball are gaining speed and so it is heating up, when the solid's temperature reaches the melting point, it will become a liquid.

Therefore, D is the correct answer.

8 0

3 years ago