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3 years ago

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A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-

dark-bright fringe shifts are observed if mirror M2 is moved 1.7000 cm ? Express your answer using five significant figures.

1 answer:

kati45 [8]3 years ago

3 0

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

$\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}$

Here,

$\Delta L$ = is the distance moved by the mirror M

$\lambda$ is the wavelenght of the light used.

$\Delta L$ = 0.017m

$\lambda = 656.45*10^{-9}m$

$\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}$

$\Delta m = 51793.72$

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

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A 121-cm-long, 4.00 g string oscillates in its m = 3 mode with a frequency of 180 Hz and a maximum amplitude of 5.00 mm. What ar

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Answer:

0.8067 m

69.696 N

Explanation:

m = Mode = 3

M = Mass of string = 4 g

f = Frequency = 180 Hz

l = Length of string = 121 cm

Length of the string is given by

$l=m\dfrac{\lambda}{2}\\\Rightarrow \lambda=2\dfrac{l}{m}\\\Rightarrow \lambda=2\times \dfrac{1.21}{3}\\\Rightarrow \lambda=0.8067\ m$

The wavelength is 0.8067 m

Linear density is given by

$\mu=\dfrac{M}{l}\\\Rightarrow \mu=\dfrac{4\times 10^{-3}}{1.21}$

Speed of the wave

$v=f\lambda\\\Rightarrow v=180\times 2\times \dfrac{1.21}{3}\\\Rightarrow v=145.2\ m/s$

Speed of wave is given by

$v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow T=\mu v^2\\\Rightarrow T=\dfrac{4\times 10^{-3}}{1.21}\times 145.2^2\\\Rightarrow T=69.696\ N$

Tension is given by 69.696 N

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A uniformly charged, straight filament 4.95 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder

Strike441 [17]

Answer:

The electric field at the cylinder surface = 80.19 kN/C

Electric flux via the cylinder $\mathbf{\phi_E = 828.63 Nm^2/C}$

Explanation:

Given that:

For the filament

The length = 4.95 m

The charge = 2.00 µC

The charge per unit length for the filament can be computed as:

$\lambda = \dfrac{q}{l}$

$\lambda = \dfrac{2}{4.5}\mu C/m$

Using Gauss's law:

$\phi_E = \oint E^{\to}*dA^{\to}$ ---- (1)

where;

electric flux = $\phi_E$

permittivity of free space = $\varepsilon_o$

electric field = E

surface area = dA

However, the electric flux $\phi_E$ via the cylinder can be expressed as:

$\phi_E = \dfrac{\lambda l'}{\varepsilon_o}$

Equation (1) can now be rewritten as:

$\dfrac{\lambda l'}{\varepsilon_o} = \oint E^{\to}*d(2 \pi rl')$

$|E| =(\dfrac{\lambda l'}{\varepsilon_o 2 \pi rl' })$

replacing the values into the above equation:

$|E| =(\dfrac{(\dfrac{2}{4.5} \mu C/m) (1.65 \ cm)}{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (10 \ cm )(1.65 cm) })$

$|E| =(\dfrac{(\dfrac{2}{4.5} \times 10^{-6} C/m) }{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (0.1 \ m ) })$

$\mathbf{|E| =80.19 \ kN/C}$

Thus, the electric field at the cylinder surface = 80.19 kN/C

The electric flux now is calculated using the said formula:

$\phi_E = \dfrac{\lambda l'}{\varepsilon_o}$

$\phi_E = \dfrac{(\dfrac{2}{4.5}\times 10^{-6} \ C)(0.0165 \ m)}{8.85 \times 10^{-12} \ C^2/Nm^2}$

$\mathbf{\phi_E = 828.63 Nm^2/C}$

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3 years ago

Walk 42 miles due north, deviate 78 degrees to east, and walk 65miles. What is the displacement? ( magnitude and direction with

pochemuha

Answer:84.405m, $\theta =48.876^{\circ}$

Explanation:

Given

Person walk 42 miles due to north so its position vector is

$r_1=42\hat{j}$

Now he deviates $78^{\circ}$ to east and walk 65 miles

so its new position vector

$r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}$

$r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}$

So magnitude of acceleration is

$|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}$

$|r_2|=\sqrt{63.58^2+55.514^2}$

$|r_2|=84.405 m$

for direction

$tan\theta =\frac{42+65cos78}{65sin78}$

$tan\theta =0.8731$

$\theta =41.124^{\circ}with\ respect\ to\ east$

$\theta =48.876^{\circ}with\ respect\ to\ North$

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