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timofeeve [1]
3 years ago
15

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

each a point on a far-away screen such thatrays from the slit make an angle of 1.0° with the normal. Thedifference in phase for waves from the top and bottom of the slitis:
A) 0
B) 0.55 rad
C) 1.1 rad
D) 1.6 rad
E) 2.2 rad
Physics
1 answer:
kaheart [24]3 years ago
6 0

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

\Phi = \frac{2\pi \delta}{\lambda}

Where

\delta = Horizontal distance between two points

\lambda = Wavelength

From our values we have,

\lambda = 500nm = 5*10^{-6}m

\theta = 1\°

The horizontal distance between this two points would be given for

\delta = dsin\theta

Therefore using the equation we have

\Phi = \frac{2\pi \delta}{\lambda}

\Phi = \frac{2\pi(dsin\theta)}{\lambda}

\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}

\Phi= 1.096 rad \approx = 1.1 rad

Therefore the correct answer is C.

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Which of these statements about a dipole are correct? Select all that are true.
krok68 [10]

Answer:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

Explanation:

A dipole is a pair of magnetized, equal or oppositely charged poles the are being separated by a distance.

The statements about a dipole that are correct are:

• The electric field at any location in space, due to a dipole, is the vector sum of the electric field due to the positive charge and the electric field due to the negative charge.

• At a distance d from a dipole, where d >> 5 (the separation between the charges), the magnitude of the electric field due to the dipole is proportional to 1/d^3

• A dipole consists of two particles whose charges are equal in magnitude but opposite in sign

7 0
2 years ago
Two compact sources of sound oscillate in phase with a frequency of 100 Hz. At a point 5.00m from one source and 5.85 m from the
antoniya [11.8K]

Answer:

(a)1.557 radian (b) 1.424 A

Explanation:

Frequency of oscillation of sound = 100 Hz

\Delta r=5.85-5=0.85m

(a)The phase difference is given by \frac{2\pi \Delta r}{λ} where v is the velocity of sound in air

So phase difference \frac{2\pi \Delta rf}{v} as \lambda =\frac{v}{f}

So phase difference =\frac{2\times 3.14\times 0.85\times 100}{343}=1.557radian

(b) The  resultant amplitude is given by 2Acos\frac{\Phi }{2}=2\times A\times cos\frac{1.557}{2}=1.424A

5 0
2 years ago
A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a rampThe average power deve
Flauer [41]

Answer:

there it is fella

we neglect the mass data

3 0
3 years ago
Can someone please help
ki77a [65]

Answer:

Acceleration of that planet is 30 \frac{m}{s^{2} }.

Given:

initial speed of hammer = 0 \frac{m}{s}

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + \frac{1}{2} g 1^{2}

g = 15 × 2

g = 30 \frac{m}{s^{2} }

Thus, acceleration of that planet is 30 \frac{m}{s^{2} }.

8 0
3 years ago
Which graph uses bars to show data that are broken into intervals?
AnnyKZ [126]

Answer:

A. scatter plot?

Explanation:

I dont really know if I'm right... sorry.

5 0
2 years ago
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