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Ira Lisetskai [31]
3 years ago
8

Suppose you see two main-sequence stars of the exact same spectral type. Star 1 is dimmer in apparent brightness than Star 2 by

a factor of 100. What can you conclude? (Neglect any effects that might be caused by interstellar dust and gas.)
Physics
1 answer:
katrin2010 [14]3 years ago
3 0

Options:

A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.

B. Star 1 is 100 times more distant than Star 2.

C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.

D. Star 1 is 10 times more distant than Star 2.

E. Star 1 is 100 times nearer than Star 2.

Answer:

D. Star 1 is 10 times more distant than star 2

Explanation:

For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.

Luminosity, L = 1/r²

Where r is the distance of the star to the earth

Since star 1 is dimmer in brightness than star 2 by a factor of 100,

L₁/L₂ = 1/100

i.e. L₁ = 1, L₂=100

L₁ = 1/r₁² ............(1)

1 =  1/r₁²

L₂ = 1/r₂²

100 =  1/r₂² .........(2)

divide equation (2) by equation (1)

100/1 = ( 1/r₂² )/ (1/r₁²)

100 = (r₁/r₂)²

r₁/r₂ = √100

r₁/r₂ = 10

r₁ = 10r₂

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A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered
mart [117]

Answer:

\omega_f=12.2954 \,rad.s^{-1}         i.e.

Frequency=1.9569\,rev.s^{-1}

Explanation:

Given:

angular speed,\omega_i =2\times 2\pi \,rad.s^{-1}

mass of the disk, M=4.9\,kg

radius of the disk, R=0.2\,m

mass of the chunk, m=2.7\,kg

radius of the chunk, r=0.04\,m

We know that the angular momentum is given by:

L=I.\omega

and moment of inertia for a disc:

I=\frac{1}{2} m.r^2

According to the conservation of angular momentum, the final angular momentum is equal to the initial angular momentum.

\frac{1}{2} \times M.R^2\times \omega= \frac{1}{2} \times (M.R^2+m.r^2) \omega_f

4.9\times 0.2^2\times (2\times 2\pi)=(4.9\times 0.2^2+2.7\times 0.04^2 )\tiems \omega_f

\omega_f=12.2954 \,rad.s^{-1}

Frequency=1.9569\,rev.s^{-1}

4 0
4 years ago
Which wave has a disturbance that is parallel to the wave motion?
Leto [7]
The answer is D.<span>longitudinal</span>
6 0
3 years ago
Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

5 0
3 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
Which of the following changes will increase the frequency of an oscillating pendulum?
svlad2 [7]

Answer:

c. an increase in the length of the rope.

6 0
3 years ago
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