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Andreas93 [3]
3 years ago
14

compare to visible light, the wavelength of x-rays is shorter longer or the same and the frequency is lower higher or the same

Physics
1 answer:
Doss [256]3 years ago
7 0

Answer:shorter higher

Explanation:Compare to visible light, the wavelength of X-rays is shorter and then frequency is higher

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The big bang theory is the most accepted theory regarding the origin of the solar system. It suggests that our star, the Sun, was first created by a cloud of<span>dust and gas.</span>
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Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface
Goryan [66]

Answer:

so maximum velocity for walk on the surface of europa is  0.950999 m/s

Explanation:

Given data

legs of length r =  0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr)    ...............1

here r is length of leg

we know g = GM/r²   from universal gravitational law

so G we know 6.67 × 10^{-11} N-m²/kg²

g = 6.67 × 10^{-11} ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

we put all value in equation 1

max velocity = √(1.33 × 0.68)

max velocity = 0.950999 m/s

so maximum velocity for walk on the surface of europa is  0.950999 m/s

3 0
3 years ago
Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2
igomit [66]

Answer

Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

      \lambda_{max} = 0.5 \mu_m

2) for tungsten T = 2500 K

      \lambda_{max} = \dfrac{2898}{2500}

      \lambda_{max} = 1.16 \mu_m

3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

      \lambda_{max} = \dfrac{2898}{60}

      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

8 0
3 years ago
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sashaice [31]

Answer:

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Explanation:

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siniylev [52]
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I hope this helped ! Sorry if it’s wrong :)
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