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mars1129 [50]
3 years ago
15

You are doing x-ray diffraction on a crystal that has a cubic structure, using 0.340-nm x rays. Part A If the lattice spacing is

d = 6.70 × 10−10 m, what is the greatest Bragg angle at which you observe peaks in the intensity of the diffracted rays?
Physics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

∅ = 0.26°

Explanation:

Bragg's law states that:

n×λ = 2×d×sin(∅)

<em>where:</em>

  • <em>d is the lattice spacing.</em>
  • <em>λ is the wavelength of the rays.</em>
  • <em>n is the order of the diffracted rays, n = 1.</em>

then:

sin(∅) = λ/2×d

    ∅ = sin^-1(λ/2×d)

    ∅ = 0.26°

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A 65.8-kg person throws a 0.0413 kg snowball forward with a ground speed of 32.5 m/s. A second person, with a mass of 58.7 kg, c
guapka [62]

Answer:

v_{1} = 2.490\,\frac{m}{s}

v_{2} = 0.023\,\frac{m}{s}

Explanation:

The statement is described physically by means of the Principle of Momentum Conservation. Let assume that first person moves in the positive direction:

First Person

(65.8\,kg)\cdot (2.51\,\frac{m}{s}) = (65.8\,kg)\cdot v_{1} + (0.0413\,kg)\cdot (32.5\,\frac{m}{s} )

Second Person

(0.0413\,kg)\cdot (32.5\,\frac{m}{s})+(58.7\,kg)\cdot (0\,\frac{m}{s})=(0.0413\,kg+58.7\,kg)\cdot v_{2}

The final velocities of the two people after the snowball is exchanged is:

v_{1} = 2.490\,\frac{m}{s}

v_{2} = 0.023\,\frac{m}{s}

6 0
3 years ago
Railroad car A, with mass 6275 kg, is traveling east at 6.5 m/s along a straight track. It strikes railroad car B, with mass 515
LekaFEV [45]

The speed of A and B immediately after collision is 5.28m/s

<u>Explanation:</u>

Mass of A is 6275kg

Speed of A is 6.5m/s

Mass of B is 5155kg

Speed of B is 3.8m/s

Track is frictionless.

A and B stick together.

speed of attached A and B = ?

mₐsₐ + mᵇsᵇ = (mₐ + mb) s

6275 X 6.5 + 5155 X 3.8 = ( 6275 + 5155) X s\\\\s = \frac{40787.5 + 19589}{11430}\\ \\s = \frac{60376.5}{11430}\\ \\s = 5.28m/s

Therefore, The speed of A and B immediately after collision is 5.28m/s

4 0
3 years ago
Which statement best describes the situation? Which statement best describes the situation? The car has the greater change of mo
Pachacha [2.7K]

Complete Question

A truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation?

A. the truck has the greater change of momentum because it has the greater mass

B. the car has the greater change of momentum because it has the greater speed

C. neither the car nor the truck changes its momentum in the collision because momentum is conserved

D. they both have the same change in magnitude of momentum because momentum is conserved

E. none of the above is necessarily true

Answer:

D. They both have the same change in magnitude of momentum because     momentum is conserved

Explanation:

In order to get a good understanding of the solution above we define some

concetps

   Momentum

This is defines quantified motion and can be mathematically represented as  

      Momentum =  Mass of the body × Velocity of the body

According to the Law of conservation of momentum states that when two particles collide together in a system that is being isolated that their total momentum before and after their collision is equal this means that the momentum lost by the truck would be the same as the momentum gained by the small car

4 0
3 years ago
A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui
GalinKa [24]

Answer:

n_l = 1.97

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

focal length of lens in liquid is

\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

                =\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}

rearrange fron_l

n_l = \frac{n_g f_l}{f_l+f(n_g-1)}

n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}

n_l = 1.97

7 0
3 years ago
One mole of an unknown sample contains 120 g of carbon and 30.3 g of
barxatty [35]

atomic mass of carbon is C = 12 g/mol

atomic mass of hydrogen is H = 1 g/mol

now number of carbon atoms

N_c = \frac{120}{12} = 10

Similarly the number of hydrogen atoms are

N_H = \frac{30.3}{1} = 30

so Carbon atom and hydrogen atom must be in ratio of 10:30

so the ratio is 1:3

now the empirical formula is always in simplest form

CH_3

5 0
2 years ago
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