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skelet666 [1.2K]
3 years ago
11

A 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 n

m. 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:
maksim [4K]3 years ago
5 0

Answer:

A) receding from the earth

B) 3.078x10^6m/s

Explanation:

  • A) receding from the earth

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

  • B) 3.078x10^6m/s

to calculate the relative speed we use the following formula:

v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )

where c is the speed of light: c=3x10^8m/s

\lambda_{1} is the wavelength emited by the source, and

\lambda_{2} is the wavelength measured on earth.

we substitute all the values and do the calculations:

v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s

the relative speed is: 3.078x10^6m/s

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Answer

t = 367.77 s = 6.13 min

Explanation:

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where,

P = Electric Power of Heater =  300 W

t = time required = ?

m_g = mass of glass = 300 g = 0.3 kg

m_w = mass of water = 250 g = 0.25 kg

C_g = speicific heat of glass = 840 J/kg.°C

C_w =  specific heatof water = 4184 J/kg.°C

ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C

ΔT_g = ΔT_w = 85°C

Therefore,

(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\

<u>t = 367.77 s = 6.13 min</u>

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3 years ago
After creating a prototype, which of the following would not be and appropriate next step in the design process of a new technol
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The answer would be to research the need.  This should have been done before the project began.

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A tort that permanently interferes with another's use and enjoyment of his or her personal property.

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v = λ / T

λ = v * T

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3 years ago
The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the ma
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Answer:

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Total acceleration is the vector sum of  tangential acceleration and radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\

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a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2

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a_c = \frac{v^2}{r}

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