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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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A car of mass 1500 kg travels due East with a constant speed of 25.0 m/s. Eventually it turns right, and travels due South with
e-lub [12.9K]

Answer:

The direction of the car’s change in linear momentum is 149.04° West of North

Explanation:

Momentum is defined as the product of mass of a body and its velocity

Momentum = mass × velocity

Change in Momentum = mass × change in velocity

∆P = m∆v

∆P = m(v-u)

Given m = 1500kg

v = 25m/s

u = 15m/s

∆P = 1500(25-15)

∆P = 1500×10

∆P = 15,000kgm/s

Since the car first travels due East i.e +x direction

x = 25m/s

Travelling due south is negative y direction

y = -15m/s

Direction of the car change

θ = tan^-1(y/x)

θ = tan^-1(-15/25)

θ = tan^-1(-0.6)

θ = -30.96°

Since tan is negative in the second quadrant

θ = 180-30.96

θ = 149.04°

The direction of the car’s change in linear momentum is 149.04° West of North

5 0
4 years ago
A block sliding across a level surface has a mass of 2.5 kg and a mechanical energy of 20 joules. What is its velocity?
tia_tia [17]
Hi!

The energy of the block is 4 m/s

To calculate this, you need to use the equation for kinetic energy. The block is sliding (i.e. it's moving). If the object is sliding across a level surface, the only energy it has is kinetic energy, because there is no change in potential energy (which changes with height). So, the mechanical energy will be pure kinetic energy. The equation is the following, derived from the expression for kinetic energy:

v= \sqrt{ \frac{2*Ke}{m}}=\sqrt{ \frac{2*20 (kg*m^{2}*s^{-2}) }{2,5kg}}=4 m/s

Have a nice day!
8 0
3 years ago
What does the energy hill represent on an energy diagram?
creativ13 [48]

Answer:

In my opinion I think the answer is C you don't have to choose C

8 0
3 years ago
How much work did the movers do (horizontally) pushing a 43.0-kg crate 10.4 m across a rough floor without acceleration, if the
Ilia_Sergeevich [38]

The crate is in equilibrium. Newton's second law gives

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

where

• <em>n</em> = magnitude of the normal force

• <em>mg</em> = weight of the crate

• <em>p</em> = mag. of push exerted by movers

• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>

<em />

It follows that

<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N

so that the movers perform

<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J

of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)

8 0
3 years ago
Why do the earth have many states
IrinaVladis [17]
Landforms pushing up mountains and hills. water and wind can wear down land and create valleys and canyons
6 0
3 years ago
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