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sergiy2304 [10]
3 years ago
6

An x-ray beam with wavelength 0.240 nm is directed at a crystal. as the angle of incidence increases, you observe the first stro

ng interference maximum at an angle 24.5 ∘. what is the spacing d between the planes of the crystal?
Physics
1 answer:
jeka943 years ago
5 0
This phenomenon is called the Bragg scattering. The working equation used for this is:

λ = 2dsinθ
where
λ is the wavelength and θ is the incident angle

0.240nm = 2dsin(24.5°)
Solving for d,
<em>d = 0.29 nm</em>
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lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of
bonufazy [111]

Answer:

The intensity of laser 2 is 4 times of the intensity of laser 1.

Explanation:

The intensity in terms of electric field is given by :

U=\dfrac{1}{2}\epsilon_o E^2

E is electric field

It means, U\propto E^2

In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.

Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

\dfrac{U}{U'}=(\dfrac{E}{E'})^2

We have,

E'=2E

So,

\dfrac{U}{U'}=\dfrac{E^2}{(2E)^2}\\\\\dfrac{U}{U'}=\dfrac{E^2}{4E^2}\\\\\dfrac{U}{U'}=\dfrac{1}{4}\\\\U'=4\times U

So, the intensity of laser 2 is 4 times of the intensity of laser 1.

6 0
3 years ago
A pure sound wave, generated by a tuning fork, is considered a periodic wave. Which statement is true for this tuning fork sound
soldier1979 [14.2K]

Answer : Every wave has the same wave pattern.

Explanation : A pure sound wave has a single frequency and its generated by a tuning fork . The sound wave is a  simple periodic wave.

When an object such as a guitar push and pull on the around it. When it pushes on the air, then the pressure increases and when it pull on the air,  then the pressure decreases and sound waves are formed.

Hence, sound wave has the same wave pattern.  

4 0
3 years ago
Read 2 more answers
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
If you use the same force to push a motorcycle as you would push a bike which one would have more acceleration and why explain u
zaharov [31]

Answer:

The bike would have more acceleration

Explanation:

Accourding to newtons first law a force is equal to its mass multiplied by its acceleration (f=ma) therefore an object with a higher mass compared to an object with a lower mass would experience less acceleration.

Eg.

F=50N

Motorbike M=200kg

F=ma

50=200 x a

50/200=a

0.25m/s/s =a

Bike M=35kg

F=ma

50=35 x a

50/35= a

1.43m/s/s=a

6 0
3 years ago
Read 2 more answers
A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c
Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; m_{totalv

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg +  0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

3 0
3 years ago
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