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USPshnik [31]
3 years ago
15

distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.

(a) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. (b) Find the speed of the galaxy relative to the earth.
Physics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer:

The speed of the galaxy relative to the Earth is 3.09\times 10^6\ m/s.

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s

So, the speed of the galaxy relative to the Earth is 3.09\times 10^6\ m/s.

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Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
RideAnS [48]

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

A_{1}\times V_{1} = A_{2}\times V_{2}

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

A_{1} = 0.070 m^{2}

V_{1} = 3.50 m/s

a) A_{2} = 0.105 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.105\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.150}m/s

or

V_{2} = 2.33m/s

b) A_{2} = 0.047 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.047\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.047}m/s

or

V_{2} = 5.21m/s

c) we have,

DischargeQ = Area (A)\times Velocity(V)

thus from the given value, we get

Q = 0.070m^{2}\times 3.5m/s\

Q = 0.245 m^{3}/s

Also,

DischargeQ = \frac{volume}{time}

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

0.245m^{3}/s = \frac{volume}{3600s}

or

0.245m^{3}/s\times 3600 = Volume

volume of flow = 882 m^{3}

8 0
3 years ago
Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation, in kW, from 1
Leya [2.2K]

Answer:

1.5 x 10⁵ W

Explanation:

A = Area of the fresh lava = 1.02 m²

T = Temperature of fresh lava = 1000 °C = 1000 + 273 = 1273 K

T₀ = Temperature of surrounding = 25.3 °C = 25.3 + 273 = 298.3 K

ε = emissivity of the lava = 0.97

σ = stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Rate of  transfer is given as

E = σ ε A (T⁴ - T₀⁴)

E = (5.67 x 10⁻⁸) (0.97) (1.02) ((1273)⁴ - (298.3)⁴)

E = 1.5 x 10⁵ W

3 0
3 years ago
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