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Rashid [163]
3 years ago
12

Two slits in an opaque barrier each have a width of 0.020 mm and are separated by 0.050 mm. When coherent monochromatic light pa

sses through the slits the number of interference maxima within the central diffraction maximum: Group of answer choices is 1 is 4 is 2 is 5 cannot be determined unless the wavelength is given
Physics
1 answer:
Semenov [28]3 years ago
5 0

Answer:

The answer is 5

Explanation:

The maximum interference is:

m * λ = d * sinθi

Where m = 0,1,2,3,...

The first minimum diffraction is:

λ = a * sinθd

|sinθi| < sinθd

Where

(|m| * λ)/d < λ/a

|m| < d/a = 2.5

|m|max = 2

It can be concluded that coherent monochromatic light passes through the slits, therefore the maximum number of interference is 5.

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B is correct. Ever wondered why ice floats in water?
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Two identical cars are parked in the sun on a warm day. One car is black and the other is white. Both cars are parked for three
cestrela7 [59]

Answer:

black will be hotter because black paint absorbs light and white reflects it.

4 0
2 years ago
Read 2 more answers
Please help need this asap
NikAS [45]

Answer:

A)  350 N

B)  58.33 N

C)  35 kg

D)  35 kg

Explanation:

If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2

The weight of the object on Earth is given by:

Weight = mass * g = 35 * 10 = 350 N

The weight of the object on the Moon:

Weight = mass * gmoon = 35 * 5/3 = 58.33 N

The mass of the object on Earth is 35 kg

The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.

3 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
2 years ago
Why is the air drag on a baseball different than it would be for a smooth ball with no stitches? How does this apply to the desi
yKpoI14uk [10]

Answer:

The stitches and dimples around a baseball  and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.

Explanation:

The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.  

A smooth ball with no stitches or dimples has more air drag that opposes the motion.

A golf ball is smooth ball with dimples to create a thin turbulent boundary layer of air that clings to the ball's surface. This allows the smoothly flowing air to follow the ball's surface a little farther around the back side of the ball, thereby decreasing the size of the wake, and allowing the ball to travel more easily.

7 0
3 years ago
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