What is the surface temperature of a star that has a peak wavelength of 290 nm?

1 answer:

5 0

We need to use Wien's Law

Wavelength = 0.0028976 [m.K] / T

This establishes a relation between the wavelength and temperature of a black body (any body that absorbs radiation, such as the stars)

T = 0.0028976 [m.K]/290 E-9[m] = 9991.724 K

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200 kW of solar radiation is shining on a 300 m^2 parking lot. What is the insulation on the parking lot?

ale4655 [162]

That's "insolation" ... not "insulation".

'Insolation' is simply the intensity of solar radiation over some area.

If 200 kW of radiation is shining on 300 m² of area, then the insolation is

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Note that this is the intensity of the incident radiation. It doesn't say anything
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Wait !
I just looked back at the choices, and realized that I didn't answer the question
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Wait again !
I found it, through literally several seconds of online research.

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3 years ago

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raketka [301]

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2 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$