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melomori [17]
2 years ago
12

What is the frequency of radiation whose wavelength is 11.5 a0 ?

Physics
1 answer:
irakobra [83]2 years ago
4 0

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

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1 BTU mean 1 degree of F increased for 1 pound mass. First, we can calculate the mass of the water in lb unit. The calculation would be:

mass= energy/temperature increase = 500,000/80= 6250lb.

Then we need to convert the weight into gallons. The calculation would be:
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7 0
2 years ago
A 3.0-cm tall statue is 48 cm in front of a convex lens having a focal length of i. (2 pts) Is the image of the statue real or v
erica [24]

Explanation:

Given that,

Height of object = 3.0 cm

Distance of object u= 48 cm

Focal length = 48 cm

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Using formula of lens

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Put the value into the formula

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{-48}

\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{48}

\dfrac{1}{v}=\dfrac{17}{240}

v=14.11\ cm

(I). The image is real.

(II).  The distance of the image from the lens is 14.11 cm.

(II). The image is inverted.

(IV). We need to calculate the height of the image

Using formula of magnification

m = \dfrac{v}{u}=\dfrac{h'}{h}

\dfrac{v}{u}=\dfrac{h'}{h}

Put the value into the formula

\dfrac{14.11}{48}=dfrac{h'}{3.0}

h'=\dfrac{14.11}{48}\times3.0

h'=0.88\ cm

The height of the image is 0.88 cm.

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6 0
2 years ago
A thin partition divides a thermally insulated vessel into a lower compartment of volume V and an upper compartment of volume 2V
Anni [7]

Answer:

1.89mol

Explanation:

The entropy change during free expansion is express as

S_{f}-S_{i}=nRln(\frac{V_{F}}{V_{I}})\\

Where S is the entropy of the system,

            n is the amount of mole

             R is the gas constant = 8.314 and

           V is the volume occupied at the initial and final stage

since the process is n adiabatic free expansion, the entropy of the system is constant. Hence we can re-write the equation as

S=nRln(\frac{V_{F}}{V_{I}})\\

where the  V_{i}=v\\ and V_{f}=2v+v=3v\\

S=17.28J/k\\ and

R=8.314\\

Now if we substitute in values we arrive at

17.28=(8.314)n*ln(\frac{3v}{v} )\\17.28=(8.314)n*ln(3 )\\17.28=(8.314)n*1.0986\\n=\frac{17.28}{8.314*1.0989}\\n=1.89 mole\\

6 0
2 years ago
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balu736 [363]

Answer:

#4 is the accurate answer.

8 0
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A 92-kg rugby player running at 7.5 m/s collides in midair with a 112-kg player moving in the opposite direction. After the coll
madam [21]

By the condition of momentum conservation we can say that since there is no external force along horizontal direction so we will have

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here we know that

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also we know that

v_{1f} = v_{2f} = 0

now from above equation we have

92(7.5) + 112 v = 0 + 0

now we have

v = 6.16 m/s

so the speed of the other player must be 6.16 m/s

8 0
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