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olasank [31]
3 years ago
8

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Engineering
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

Frequency = 3.5294\times 10^{14}s^{-1}

Wavenumber = 1.1765\times 10^6m^{-1}

Energy = 2.3365\times 10^{-19}J

Energy = 1.4579 eV

Energy = 2.3365\times 10^{-12}erg

Explanation:

As we are given the wavelength = 850 nm

conversion used : (1nm=10^{-9}m)

So, wavelength is  850\times 10^{-9}m

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where, c is the speed of light having value = 3\times 10^8m/s

So, Frequency is:

Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}

Frequency=3.5294\times 10^{14}s^{-1}

Wavenumber is the reciprocal of wavelength.  

So,  

Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}

Wavenumber=1.1765\times 10^6m^{-1}

Also,  

Energy=h\times frequency

where, h is Plank's constant having value as 6.62\times 10^{-34}J.s

So,  

Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})

Energy=2.3365\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=1.4579eV

Also,  

1J=10^7erg

So,  

Energy=(2.3365\times 10^{-19})\times 10^7erg

Energy=2.3365\times 10^{-12}erg

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The 2-kg package leaves the conveyor belt at with a speed of vA = 1 m/s and slides down the smooth ramp. Determine the required
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Answer:

Incomplete question: Find the normal reaction the curved portion of the ramp exerts on the package at B if pB = 2 m, the height is 4 m

Answer: The required speed is 8.9107 m/s

The normal reaction is 99.0006 N

Explanation:

Given data:

m = 2 kg

vA = speed = 1 m/s

h = 4 m

g = gravity = 9.8 m/s²

Questions: Determine the required speed of the conveyor, vc = ?

Find the normal reaction, Nc = ?

The required speed applying the conservation of energy:

\frac{1}{2} mv_{A}^{2}  +mgh=\frac{1}{2} mv_{c} ^{2}

Solving for vc:

v_{c} =\sqrt{v_{A}^{2}+2gh  } =\sqrt{1^{2}+(2*9.8*4) } =8.9107m/s

The normal reaction applying the equilibrium of forces:

N_{c} -mg=m(\frac{v_{c}^{2}  }{p_{B} } )

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3 years ago
A cylindrical shell of inner and outer radii, ri and ro, respectively, is filled with a heat-generating material that provides a
mestny [16]

Answer:

A)The expression for the steady-state temperature distribution T(r) in the shell is;

T(r) = [(q'r²/4k)(ro² - r²)] + [(q'ri/2k)In(r/ro)] + (q'ro/2h)[1 - (ri/ro)²] + T∞

B) The expression for the heat flux, q''(ro), at the outer radius of the shell is; q'(ro) = q'π(ro² - ri²)

Explanation:

A) we want to find an expression for the steady-state temperature distribution T(r) in the shell.

First of all, one dimensional radial heat for cylindrical shell with uniform heat generation is;

(1/r)(d/dr)[rdT/dr] + (q'/k) = 0

Subtract (q'/k)from both sides to give;

(1/r)(d/dr)[rdT/dr] = - (q'/k)

Multiply both sides by r to give;

(d/dr)[rdT/dr] = - (q'r/k)

So, [rdT/dr] = - ∫(q'/k)

So [rdT/dr] = -(q'r²/2k) +C1

And;

[dT/dr] = - (q'r²/2k) +(C1)/r

Thus, T(r) = - (q'r²/4k) +(C1)In(r) +C2

Now, we apply the boundary condition at r = ri for dT/dr =0

Thus; (dT/dr)| at r=ri, is zero.

Thus, at r=ri

d/dr[(q'r²/2k) +(C1)In(r) +C2] = 0

Thus;

- (q'ri/2k) +(C1)/ri + 0 = 0

So, making C1 the subject of the formula,

C1 = (q'ri/2k)

Now, let's apply the boundary condition at r=ro for

q''(conduction) = q''(convection)

So, at r=ro, - kdT = h[T(ro - T∞)]

So, using previously gotten equation above, we obtain,

-k[(q'ro²/2k) + (C1)/ro] = h[-(q'ro²/4k) +(C1)In(ro) + C2- T∞)]

So making C2 the subject, we have;

C2 = (q'ro/2h)[1 - (ri/ro)²] + (q'ro²/2k)[(1/2) - (ri/ro)²In(ro)] + T∞

So putting the formulas for C1 and C2 in the equation earlier derived for T(r) to obtain;

T(r) = - (q'r²/4k) + (q'r²/2k)In(r) + (q'ro/2h)[1 - (ri/ro)²] + (q'ro²/2k)[(1/2) - (ri/ro)²In(ro)] + T∞

Thus;

T(r) = [(q'r²/4k)(ro² - r²)] + [(q'ri/2k)In(r/ro)] + (q'ro/2h)[1 - (ri/ro)²] + T∞

B) We want to find out heat rate at outer radius of the shell. So at r=ro, Formula is;

q'(ro) = - k(2πro) dT/dr

= - k(2πro) [- (q'ro²/2k) +(C1)/ro]

C1 = (q'ri/2k) from equation earlier. Thus;

q'(ro) = -k(2πro) [- (q'ro²/2k) +(q'ri/2k)/ro]

When we expand this, we obtain;

q'(ro) = q'π(ro² - ri²)

8 0
3 years ago
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