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Mama L [17]
3 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]3 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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A 10n falling object encounters 4n of air resistance. what is the net force on the object?
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What is the mass of an object that has an acceleration of 2.63 m/s^2 when an unbalanced force of 112 N is applied to it?
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6 0
3 years ago
14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar
Zolol [24]

Answer:

T= 11.0003s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally  the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

     \omega ^2 r =Angular\ acc

Considering the outer ring,

 \omega ^2 r = 9.8

  \omega ^2= \frac{9.8}{30}

 \omega = \sqrt{\frac{9.8}{30}}

 \omega= 0.571 rad/s

Therefore solving for  Period T

Generally the equation for solving Period T is mathematically given as

 T= \frac{2\pi}{\omega}

 T= \frac{2\pi}{0.571 rad/s}

 T= 11.0003s

5 0
3 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
natima [27]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

5 0
3 years ago
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