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Mama L [17]
4 years ago
8

A scientist is studying the light emitted by several celestial objects. He records the shifts in each object’s spectral lines (i

n nm). Based on these measurements, determine whether each object is moving toward or away from Earth. Use the diagram of the visible light spectrum to help you. 400 nm shifted to 430 nm 610 nm shifted to 580 nm 512 nm shifted to 480 nm 670 nm shifted to 690 nm
Physics
1 answer:
forsale [732]4 years ago
5 0

-- 400 nm shifted to 430 nm . . . longer than it should be; "red shifted"; moving away from Earth  

-- 610 nm shifted to 580 nm . . . shorter than at source; "blue shifted"; moving toward Earth

-- 512 nm shifted to 480 nm . . . shorter than at source; moving toward Earth

-- 670 nm shifted to 690 nm . . .longer than at source; moving away from Earth

Now I'd just like to ask one more itty bitty question, that you can think about while you're on this subject:  Astronomers really do this.  They measure how much the wavelength CHANGED, from the time it left the original source until the time they detect it. But HOW do they know what the wavelength WAS when it left the source ? ? ?

THIS is the part that blows my mind !

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An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.
UNO [17]

Explanation:

The given data is as follows.

        Electric field strength (E) = 187,500 V/m

    Magnetic field strength (B) = 0.125 T

       Diameter (d) = 25.05 cm = 0.2505 m    (as 1 m = 100 cm)

    Radius (r) = \frac{d}{2}

                    = \frac{0.2505}{2}

                    = 0.12525 m

Formula to calculate the magnetic force (F_{M}) is as follows.

              F_{M} = Bqv ............ (1)

Electrical force is calculated as follows.

             F_{E} = qE ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

       F_{M} - F_{E} = 0

or,             F_{M} = F_{E}

Hence, from equations (1) and (2)

              Bqv = qE

or,            v = \frac{E}{B} ............. (3)

                  = \frac{187500 V/m}{0.125 T}

                  = 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

              F_{c} = \frac{mv^{2}}{r} ........... (4)

where,    F_{c} = centripetal force

             F_{M} = F_{c}

Using equation (1) and (4) as follows.

            F_{M} = F_{c}

              Bqv = \frac{mv^{2}}{r}

                   \frac{q}{m} = \frac{v}{Br}

                       = \frac{15 \times 10^{5}}{0.125 \times 0.12525}

                       = 958.08 \times 10^{5} C/kg

Thus, we can conclude that charge-to-mass ratio of the given particle is 958.08 \times 10^{5} C/kg.

8 0
3 years ago
Homework B5
nydimaria [60]

Answer:

I hope this helps a little bit.

7 0
3 years ago
A 730-N person pulls a rope tied to a tree. If the person pulls the rope with a force of 120 N, what is the magnitude of the rea
RUDIKE [14]

Answer: 120N

Explanation:

8 0
3 years ago
True or false forces in a force pair never cancel each other out
pishuonlain [190]
The answer is false
3 0
3 years ago
A train accelerates from a station with a = 1.841m/s ? Upon reaching a speed of 23.52m/s the train travels at a constant velocit
scZoUnD [109]

Answer:

63.29s

Explanation:

Firstly calculate the time taken to reach 23.52m/s

;use the formula...v = u + at

23.52 = 0 + 1.841t

then obtain...t = 12.78s

Then calculate the time for the last part of the journey...where the train slows down...

use the formula that is above...

0 = 23.52 - 2t...(negative for deceleration)

then obtain....t = 11.76s

Then we know that the total area under the graph of u against t..is equal to 1200m

For the first triangle(first part of the journey...where the train accelerates)

(23.52 × 12.78)÷2 = 150.3m

Then for the constant velocity part...a rectangle...

23.52 × f.....where f represents the time taken by the train having constant velocity.

...= 23.52fm

Then for the last part of the journey...the deceleration part..a triangle

(23.52 × 11.76)÷ 2 = 138.3m

Then....we add all the obtained distances and equate to 1200m so that we can obtain time (f)

138.3 + 150.3 + 23.52f = 1200

where f = 38.75s

Then total time for the whole journey of the train...

38.75 + 11.76 + 12.78

;Ans = 63.29s

3 0
3 years ago
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