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sesenic [268]
2 years ago
6

An experiment is conducted in which red light is diffracted through a single slit. Listed below are alterations made, one at a t

ime, to the original experiment, and the experiment is repeated. After each alteration, the experiment is returned to its original configuration.
A. The slit width is halved.
B. The distance between the slits and the screen is halved.
C. The slit width is doubled.
D. A green, rather than red, light source is used.
E. The experiment is conducted in a water-filled tank.
F. The distance between the slits and the screen is doubled.
Which of these alterations decreases the angles at which the diffraction minima appear?
Physics
1 answer:
Xelga [282]2 years ago
6 0

Answer:

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

Explanation:

As we know that the position of first minimum is given as

a sin\theta = N\lambda

so we have

\theta = sin^{-1}(\fracN\lambda}{a})

so width of minimum is given as

w = L\times sin^{-1}(\fracN\lambda}{a})

now if we need to decrease the angular position of minimum

1). so we can decrease the distance of screen from the slit

2). we can decrease the wavelength

3). We can increase the width of the slit

So correct answer will be

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

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A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106
defon

Answer:

4.7\cdot 10^{-14}N

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

F=qvB

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

q=1.6\cdot 10^{-19} C

v=6.5\cdot 10^6 m/s

B=4.5\cdot 10^{-2} T

So the force experienced by the electrons is

F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N

3 0
2 years ago
A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed
andriy [413]

Answer:

The final temperature is 31.94°

Explanation:

The mass of the water in the container m₁ = 200 g = 0.2 kg

The initial temperature of the water,  T₁₁ = 30°C

The mass of the iron, m₂ = 200 g = 0.2 kg

The temperature of the iron T₂₁= 50°C is immersed in the water,

The specific heat capacity of the water, c₁ = 4200 J/(kg·°C)

The specific heat capacity of the iron, c₂ = 450 J/(kg·°C)

Heat capacity relation is given by the formula;

Heat capacity Q = Mass, m × Specific heat capacity, c × Temperature change, (T₂ - T₁)

Given that energy can neither be created nor destroyed, and with the assumption that all the heat lost by the nail is gained by the water we have;

Heat lost by iron nail = Heat gained by the  water

m₁ × c₁ × (T₂ - T₁₁) = m₂ × c₂ × (T₂₁ - T₂)

Where, T₂ is the final temperature

0.2 kg × 4200 J/(kg·°C) × (T₂ - 30) = 0.2 kg × 450 J/(kg·°C) × (50° - T₂)

840·T₂ - 25200 = 4500 - 90·T₂

4500 + 25200 = 840·T₂ + 90·T₂

29700 = 930·T₂

T₂ = 29700/930 = 31.94°.

The final temperature = 31.94°.

4 0
2 years ago
Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 36.5 cm while travelin
tia_tia [17]

Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula  v=fλ

Where v= velocity of radio waves

f= frequency

λ= wavelength of wave

  • radio waves are electromagnetic waves and as such they have the speed of light which is 3*10^{8}m/s.
  • also when a wave travels from one medium to another, the wavelength changes while the frequency remains the same.
  • calculating for the frequency of the wave in air also gives us the frequency in the window glass.

f=\frac{v}{λ}

v=3*10^{8}m/s.

λ=36.5 cm = 36.5/100= 0.365m

f=\frac{3*10^{8}m/s.}{0.365m}

f=8.219*10^{8}Hz

7 0
2 years ago
A bullet is fired through a board, 14.0 cm thick, with its line of motion perpendicular to the face of the board. if it enters w
bazaltina [42]
Okay, here is my stab at this, I hope it helps!

You know the bullet's initial velocity, V₀ = 450 m/s
You know the final velocity, V = 220 m/s
You also know how long the bullet accelerates (actually decelerates), 14cm, or .14 m

With this information, you learn that you need this equation.

V² = V₀² + 2a (x - x₀), because we have all the information except a, which is the acceleration. So putting it into the equation, it looks like this.

(220m/s)² = (450m/s)² +2a(.14m - 0m)

I'll let you solve the rest, but here are some hints.  Your answer will be really big because the bullet slows down really quickly in a really small distance, and you answer will be negative, because this acceleration is causing the bullet to go slower, which is also called deceleration. Hope that helps!
4 0
2 years ago
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Which word identifies a large natural or human-made lake used to supply water?
Sidana [21]
Reservoir i think. i’m not sure
4 0
2 years ago
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