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4 years ago

Not sure on this I know it’s not D please help

Physics

1 answer:

svet-max [94.6K]4 years ago

7 0

This is the amplitude I believe.

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Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. The electromagnetic

Aleonysh [2.5K]

Answer:

The speed of the gas relative to the earth = 6.314 * $10^{6} m/s$

Explanation:

Let the observed wavelength be $\lambda_{0} = 1915 nm$

The wavelength of the source, $\lambda_{s} = 1875 nm$

Let the speed of the gas relative to the earth be = v

The observed wavelength in relation to the source wavelength, according to doppler shift in relativity can be expressed as:

$\lambda_{0} = \lambda_{s}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$

Where the speed of light, c = 3 * 10⁸m/s

$\frac{\lambda_{0}}{\lambda_{s}} = \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }\\(\frac{\lambda_{0}}{\lambda_{s}})^{2} = \frac{1-\frac{v}{c} }{1+\frac{v}{c} }\\(\frac{1915}{1875}) ^{2}( {1+\frac{v}{c}) = ({1-\frac{v}{c} }) \\$

$1.043 +\frac{1.043v}{c} = 1 - \frac{v}{c} \\\frac{1.043v}{c} + \frac{v}{c} = 1-1.043\\2.043v = -0.043c\\v = \frac{-0.043c}{2.043} \\v = -0.02105c\\v = -0.02105 * 3 * 10^{8} \\v = -6314243.76 m/s\\The absolute value, v = 6314243.76 m/s\\v = 6.314 * 10^{6} m/s$

8 0

3 years ago

What is gravitational acceleration

Thepotemich [5.8K]

Answer:

gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag). This is the steady gain in speed caused exclusively by the force of gravitational attraction.

6 0

3 years ago

Read 2 more answers

B. A police officer is chasing a purse-snatcher down a

Art [367]

Explanation:

U better start shooting

3 0

3 years ago

At sunrise two old women started to walk towards each other. one started from point a and went towards point b while the other s

konstantin123 [22]

Let the distance between points a and b = d

For the women starting from a her velocity be $V_1$ and for the women starting from b her velocity be $V_2$

Let the rising time of sun = x AM

We have $V_1$ = $\frac{d}{12+4-x}$

and we also have $V_2$ = $\frac{d}{12+9-x}$

At they meet means, they both travel a combined distance of d

So we have $V_1*(12-x)+V_2*(12-x)=d$

Substituting velocity values we will get

$\frac{d}{12+4-x}*(12-x)+\frac{d}{12+9-x}*(12-x)=d\\ \\ \frac{12-x}{12+4-x}+\frac{12-x}{12+9-x}=1\\ \\ \frac{12-x}{16-x}+\frac{12-x}{21-x}=1\\ \\ 252-21x-12x+x^2+192-12x-16x+x^2=336-16x-21x+x^2\\ \\ x^2-24x+108=0\\ \\ (x-6)(x-18)=0\\ \\$

x= 6 AM or 18 AM, but 18 AM is not possible,

Sun rise time is 6 AM

3 0

3 years ago

I need help idk if it’s C or D?

alex41 [277]

I think the answer is D.

6 0

3 years ago