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Blababa [14]
3 years ago
6

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at

point P, what should be the path difference between the two waves? Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves? The path difference between the two waves should be one-half of a wavelength. The path difference between the two waves should be one wavelength. The path difference between the two waves should be four wavelengths. The path difference between the two waves should be one and one-quarter of a wavelengths. The path difference between the two waves should be two wavelengths. The path difference between the two waves should be one-quarter of a wavelength.
Physics
1 answer:
labwork [276]3 years ago
8 0

Answer:

Explanation:

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves?

The path difference between the two waves should be one and one-quarter of a wavelengths.

The path difference between the two waves should be two wavelengths.

The path difference between the two waves should be one-half of a wavelength.

The path difference between the two waves should be one wavelength.

The path difference between the two waves should be one-quarter of a wavelength.

The path difference between the two waves should be four wavelengths

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For a wire has a circular cross section with a radius of 1.23mm.
Mila [183]

Answer:

5.731\times 10^{-5}\ m/s

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = 1.6\times 10^{-19}\ C

n = Conduction electron density in copper = 8.49\times 10^{28}\ electrons/m^3

v_d = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s

The drift speed of the electrons is 5.731\times 10^{-5}\ m/s

v_d=\dfrac{I}{neA}

From the equation we can see the following

v_d\propto \dfrac{1}{n}

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0
3 years ago
Doubling the frequency of a wave source doubles the speed of the wave assuing that wavelength remains the same? True or false? W
SVETLANKA909090 [29]

Answer:true,because velocity is directly proportional to speed or velocity

Explanation:

Velocity = frequency x wavelength

The velocity or speed varies directly with the frequency, so as the frequency is increased, the velocity or speed is also increased

8 0
3 years ago
A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3
marissa [1.9K]

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

mg(H-h)=\frac{1}{2}mv^2

g(H-h)=\frac{1}{2}v^2

\Rightarrow H=\frac{v^2}{2g}+h

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

=v\frac{2g}{h}

\Rightarrow v^2=\frac{gD^2}{2h}

thus, height H in terms of D  is given by

H=\frac{D^2}{2h}+h

H=\frac{13.9^2}{2\times3}+3

H=35.20 m

4 0
3 years ago
A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an
Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

<u>ΔP = 0.056 psi</u>

8 0
3 years ago
Plaease help me with thesse four 40 points
Zielflug [23.3K]

Answer:

For the first one c is the answer

For the second one c is also the answer

For the third one is b

Explanation:

I took that

3 0
2 years ago
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