What is the total energy equation?

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

klemol [59]

To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.

By definition the wavelength is given defined by,

$\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}$

Where

$\lambda_{obs}$ = Observed wavelength

$\lambda_s$ = Wavelength of the source

c = Speed of light in vacuum

u = Relative velocity of the source to the observer

According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is $\lambda_{obs}=1945nm$

Therefore replacing in the previous equation we have,

$1945 = 1875 \sqrt{\frac{1+\frac{u}{c} }{1-\frac{u}{c} }}$

$\sqrt{\frac{1+u/c}{1-u/c}} = 1.03733$

$\frac{1+\frac{u}{c} }{1-\frac{u}{c} }=1.03733^2$

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

Solving for u,

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

$1+\frac{u}{c} =1.03733^2-1.03733^2(\frac{u}{c} )$

$\frac{u}{c} +1.03733^2\frac{u}{c} =1.03733^2-1$

$2.88595\frac{u}{c}=1.03733^2-1$

$\frac{u}{c} = \frac{1.03733^2-1}{2.88595}$

$u = \frac{1.03733^2-1}{2.88595}*c$

$u = 0.02635c$

Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.

6 0

3 years ago

50 POINTS PLSS

Ostrovityanka [42]

Answer:I think the answer is 5.43 I don’t rlly know

Explanation:

6 0

2 years ago

What is the advantage of SI unit over CGS unit?

Allushta [10]

ADVANTAGE

PEOPLE USED CERTAIN UNIT OF MEASUREMENT IN THEIR SPECIFIC REGION WHICH IS NOT ACCEPTED BY WORLD IN CGS SYSTEM.CGS UNIT MADE

DIFFICULTY IN TRADE. LATER SI UNIT OF MEASUREMENT HAVE BEEN ADOPTED IN 1960 AD TO BRING THE uniformity IN measurement throughout THE WORLD. SI SYSTEM MADE THE MEASURENT AROUND WORLD MORE EFFECTIVE AND LESS COFUSING THAN CGS UNIT.

5 0

3 years ago

The following graph shows the force exerted on and the displacement of object being pulled

Tomtit [17]

The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.

<h3>Work done by the applied force</h3>

The area under force versus displacement graph is work done.

The total work done by pulling the object 7 m, can be grouped into two areas;

First area, A1 = area of triangle from 0 m to 2.0 m
Second area, A2 = area of trapezium, from 2.0 m to 7.0 m

A1 = ¹/₂ bh

A1 = ¹/₂ x (2) x (20)

A1 = 20 J

A2 = ¹/₂(large base + small base) x height

A2 = ¹/₂[(7 - 2) + (7-3)] x 50

A2 = ¹/₂(5 + 4) x 50

A2 = 225 J

<h3>Total work done </h3>

W = A1 + A2

W = 20 J + 225 J

W = 245 J

Learn more about work done here: brainly.com/question/8119756

3 0

1 year ago

A 10v battery is connected in series with 2 resistors. R1 is 1 ohm and R2 is 4 ohms. What is the current that goes across R1?

Digiron [165]

Answer:

Current in circuit = 2 amp

Explanation:

Given:

Voltage of battery = 10 V

First Resistance R1 = 1 ohm

Second Resistance R2 = 4 ohm

Resistor connected in series

Find:

Current in circuit

Computation;

Resistor connected in series

So,

Total resistance R = First Resistance R1 + Second Resistance R2

Total resistance R = 1 ohm + 4 ohm

Total resistance R = 5 ohm

Current in circuit = V / R

Current in circuit = 10 / 5

Current in circuit = 2 amp

7 0

2 years ago