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grandymaker [24]
4 years ago
13

A student shines a monochromatic light through a diffraction rating and observes the resulting interference pattern on a screen.

If the student wishes to increase the distance between the first order fringe and the central maxima, the student should
A. use a light with a larger wavelength

B. move the diffraction grating closer to the screen

C. increase the distance between the slits in the diffraction grating

D. more than one of the above options will work
Physics
1 answer:
Olin [163]4 years ago
6 0

Answer:

(C) increase the distance between the slits in the diffraction grating

Explanation:

This light always form constructive interference, the difference in distance is given λn.

where;

λ is the wavelength

n is the nth maxima

At center fringe, n = 0; the difference is 0

for the first maximum, n= 1, the difference is λ

for second maximum, n = 2, the difference is 2λ

for third maximum, n = 3, the difference is 3λ

From the above illustrations, the wavelength is constant, but the distances between the slits increases, resulting in a higher difference.

Therefore, Option C is the correct answer: increase the distance between the slits in the diffraction grating

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
What is the difference in light that is refracted compared to the light that is reflected?
Kruka [31]

Answer:

Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.

8 0
3 years ago
Read 2 more answers
The change of motion of a body is proportional to the applied force, and it takes place in the direction of a straight line in w
7nadin3 [17]

Answer:

Newton's Second Law of Motion  

Explanation:

According to Newton's second law of motion, the change in velocity of a body is directly proportional to the force applied on it. Velocity is a vector quantity. It measures the magnitude of the speed as well as its direction.

F = m a

where, F is the applied force, m is the mass and a is the acceleration.

It can also be expressed as:

F = \frac{dp}{dt}

where, p = mv ( momentum)

4 0
4 years ago
Read 2 more answers
Simple physics question, check the document. Should take about 3-5 minutes.
Ahat [919]

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

5 0
3 years ago
Consider a copper wire with a diameter of 0.450 mm. If you want to cut a piece of the wire so it has a resistance of 3.25 Ω, how
Arisa [49]

Answer:

30.7m

Explanation:

You can use the following equation:

ρ = R * \frac{S}{l}

Where ρ is the resistivity, R is the resistance, S is the Cross-sectional Area and l is the length of the wire.

We issolate l in the equation and solve with the data we have:

l = R * \frac{S}{p} = 3.25 Ω * \frac{0.25 *\pi * ((0.45/1000)^2}{1.68*10^-8}

l =  30.7m

5 0
3 years ago
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