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Evgen [1.6K]
3 years ago
8

Light of wavelength 550 nm falls on a slit that is 3.50 x 10^-3 mm wide. How far from the central . maximum will the first diffr

action maximum fringe be if the screen is 10.0 m away?
Physics
1 answer:
myrzilka [38]3 years ago
6 0
Λ= 550 * 10^{-9}\] m.

D = 3.50*10^-6 m.
d = 10 m. 

m = 1

2x = ?

Then you need to find x when n=1
or 
sinθ=λ/w=524x10^-9/0.0033=1.59x10^-4
≈θ for θ<<1 , so θ=1.59x10^-4 radians

Then you can find rhe distance of<span> bright fringe from the center:</span>d= θD=1.59x10^-4x10m ≈ 1.6 mm
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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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2 years ago
A spring in a pogo-stick is compressed 12 cm when a 40. kg girl stands on the stick. what is the force constant for the pogo-sti
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Answer: It is the same amount of weight as the girl is putting on the pogo stick. When you are pushing something downward then gravity will push back with the equal amount of force.

Explanation:

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2 years ago
Should nuclear energy be part of sustainable energy future? explain your answer
liraira [26]

Answer:

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3 years ago
A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
blondinia [14]

Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

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3 years ago
A person jumped off a cliff, what would happen?
Anastaziya [24]
Death would happen, hope this helped
8 0
2 years ago
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