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kkurt [141]
3 years ago
8

You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen

ts
1. Shining the green laser through grating A you see the first maximum 1 meter away from the center
2. Shining the red laser through grating B, you see the first maximum 1 meter away from the center

In both cases, the gratings are the same distance from the screen.

(a) What can you deduce about the gratings?
(b) What would you observe if you shone the green laser through grating B:?

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
b. (a) grating B has more lines/mm; (b) the first maximum less than 1 meter away from the center
c.(a) grating B has more lines/mm; (b) the first maximum more than 1 meter away from the center
d. (a) grating B has more lines/mm: (b) the first maximum 1 meter away from the center
e. (a) grating A has more lines/mm; (b) the first maximum more than 1 meter away from the center
f. (a) grating A has more lines/mm; (b) the first maximum 1 meter away from the center
Physics
1 answer:
Nikolay [14]3 years ago
5 0

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

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Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is 5 N

Tray makes an angle of \theta =12^{\circ}

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force W_p=W\cos \theta

W_p=5\times \cos 12=4.89 N

Along the Tray

W_{along}=W\sin \theta

W_{along}=5\times \sin 12=1.04 N

Thus 1.04 N is the magnitude of force that will cause Platter to slide down  

7 0
3 years ago
A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave
damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0
3 years ago
Letters A B, C and D represent locations on a bar magnet. Which location has the greatest magnetic force? (l) A (3) C ets (2) B
Elena L [17]
Answer: The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center. So it would be D.

6 0
3 years ago
a man can row about 4kmperhr in a still water.he rows the boat 2km up the stream and 2km back to his starting point in 2hr.how f
Nat2105 [25]

<u>Answer</u>:

The stream flowing  at a speed of 2.828 \mathrm{km} / \mathrm{hr}

<u>Explanation</u>:

Given:  

Distance = 2km (both in upstream and downstream)  

The speed in still water be x km/hr.  

The speed in upstream = 4-x  

Speed in downstream = 4+x  

Solution:

We know that, Speed = distance/time  

So, Time = distance/speed

Therefore,  

2=\left(\frac{2}{4-x}\right)+\left(\frac{2}{4+x}\right)

2=\frac{2(4+x)+2(4-x)}{(4-x)(4+x)}

2(4-x)(4+x)=2(4+x)+2(4-x)

2(4-x)(4+x)=2(4+x+4-x)

By cancelling 2 on both sides,

16-x^{2}=8

x^{2}=16-8=8

x=\sqrt{8}

x=2.828 \mathrm{km} / \mathrm{hr}

Result:

Thus the speed of the stream is 2.828 \mathrm{km} / \mathrm{hr}

7 0
3 years ago
radio signals emitted from and received by an airplane have her fruit frequency of 3.00 * 10 to the 12th Hertz and travel at the
Elanso [62]

here we know that the speed of the signal is same as the speed of light

so here we will have

v = 3 \times 10^8 m/s

the altitude of the airplane is given as

h = 1 \times 10^4 m

now we know that time taken by the signal to reach the control tower is given as

t = \frac{d}{v}

now it is given as

t = \frac{1 \times 10^4 }{3 \times 10^8}

t = 3.33 \times 10^{-5} s

so above is the time taken by the signal to reach the control tower

3 0
3 years ago
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