What is the wavelength of a light with a frequency of 6.98x10to the 14th power

A demented scientist creates a new temperature scale, the "Z scale." He decides to call the boiling point of nitrogen 0°Z and th

slavikrds [6]

Answer:

A) $Z = 0.577 C +112.931$

$Z = 0.577*(100) +112.931=170.631 Z$

B) $C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

C) $K = C +273.15$

$K = -22.41 +273.15 =250.739 K$

Explanation:

For this case we want to create a function like this:

$Z = a C + b$

Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.

The boiling point of nitrogen is -195,8 °C

The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

$a = \frac{z_2 -z_1}{c_2 -c_1}= \frac{1000 Z- 0Z}{1538C -(-195.8 C)}=\frac{1000 Z}{1733.8 C}=0.577 \frac{Z}{C}$

And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

$0 = 0.577 (-195.8) + b$

And if we solve for b we got:

$b = 0.577*195.8 =112.931 Z$

So then our lineal model would be:

$Z = 0.577 C +112.931$

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

$Z = 0.577*(100) +112.931=170.631 Z$

Part B

For this case we have Z =100 and we want to solve for C, so we can do this:

$Z-112.931 = 0.577 C$

$C=\frac{Z-112.931}{0.577}= \frac{100-112.931}{0.577}=-22.41 C$

Part C

For this case we know that $K = C +273.15$

And we can use the result from part B to solve for K like this:

$K = -22.41 +273.15 =250.739 K$

6 0

3 years ago

A LED light source contains a 0.5-Watts GaAs (Eg =1.43 eV) LED. Assuming that 0.12% of the electric energy is converted to emiss

Ray Of Light [21]

Answer:

Explanation:

energy emitted by source per second = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ ; h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10 x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

= 10 x 10⁻⁸ W

power required

= 10⁻⁸ + 10 x 10⁻⁸ W

= 11 x 10⁻⁸ W.

5 0

3 years ago

Particles at the very outer edge of Saturn’s A Ring are in a 7:6 orbital resonance with the moon Janus. If the orbital period of

vivado [14]

Answer:

14 hours 18 minutes.

Explanation:

ratio of number of orbits, so it completes 7 orbits in the time Janus does 6.

(16*60+41)*6/7=858 minutes or 14 hours 18 minutes

3 0

3 years ago

What are the answers to the question

Vitek1552 [10]

Answer: the link isnt loading

7 0

3 years ago

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$