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4 years ago

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Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon

laser) Express your answer using four significant figures. ν1 ν 1 = nothing s−1 Request Answer Part B 503.0 nm (wavelength of maximum solar radiation) Express your answer using four significant figures.

1 answer:

Drupady [299]4 years ago

7 0

Answer:

A. $f=6.1475*10^{14}Hz$

B. $f=5.9642*10^{14}Hz$

Explanation:

The frequency has an inversely proportional relationship with the concept of wavelength, the greater the wavelength, the lower the frequency. For electromagnetic waves, the frequency is equal to the speed of light, divided by the wavelength.

$f=\frac{c}{\lambda}$

A.

$f=\frac{3*10^8\frac{m}{s}}{488*10^{-9}m}\\\\f=6.1475*10^{14}Hz$

B.

$f=\frac{3*10^8\frac{m}{s}}{503*10^{-9}m}\\\\f=5.9642*10^{14}Hz$

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A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the

DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 $m/s^2$ (block), aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got N = mg cos θ,

f = $\mu_s N$

= $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$ .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$ ................(v)

Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

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Illustrated here are three different __________ of carbon. They vary by the number of __________.

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inysia [295]

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A = Area of plates = πr² ( where r is radius which is half of the diameter)

K = electric constant = 9×10^9

d = distance between plates = 0.5cm = 0.005 m

Let us get the area, A = πr², where r = D/2 where D = diameter

r = 32/2 = 16cm = 0.16m

A = 22/7 × (0.16)² = 0.0804 m²

By substituting this into the capacitance formula, we have that

C = 0.0804/4×3.142*9×10^9 × 0.005

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But C =Q/V where V = Ed

Hence we have that

C = Q/Ed

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142.17*10^(-12) = 24 ×10^-9 / E × 0.005

By cross multiplying

142.17*10^(-12) × E × 0.005 = 24 ×10^-9

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