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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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2 m/s

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In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init
liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,  

n  = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,  

U = − 0.061 J

The charge on an electron,  

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,  

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:  

q ₁ = n q

   = 1 ×10 ¹³ *  − 1.6 × 10 ⁻¹⁹

   = -1.6  × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6  × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and  q ₂ are the two charges.

r  is the distance between the charge and the point.

K  =  8.98755  ×  10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 =  (8.98755  ×  10 ⁹ * (-1.6  × 10⁻⁶)²) / r

r = (18.41 ×  10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0
3 years ago
A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
Maksim231197 [3]

Answer:

v=33.764m/s

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Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

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Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

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ELEN [110]

Answer: Force

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Explanation:

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