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2 years ago

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X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a

t an angle of 75.5°, what is the lattice spacing of this crystal?

1 answer:

FrozenT [24]2 years ago

6 0

Answer:

$5.1645\times 10^{-11}\ m$

Explanation:

n = Order = 1

c = Speed of light = $3\times 10^8\ m/s$

f = Frequency = $3\times 10^{18}\ Hz$

$\theta$ = Angle = $75.5^{\circ}$

Lattice spacing is given by

$d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m$

The lattice spacing of the crystal is $5.1645\times 10^{-11}\ m$

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How to not get heart attacks ? And what is some plans/list for a healthy diet ?

Mamont248 [21]

Answer:

hey there

( i )we have to Choose good nutrition. A healthy diet is one of the best weapons you have to fight cardiovascular disease. and should be physically active and stop smoking if you do these are some

life style changes though which can prevent heart attacks

( ii )An healthy diet for heart

lots of fruits and vegetables.

nuts, beans, and legumes.

whole grains.

plant-based oils, such as olive oil.

low-fat dairy products.

Explanation:

Hope it helps you

have a nice day :)

6 0

3 years ago

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

Soloha48 [4]

Answer:

The velocity of mass 2m is $v_B = 0.67 m/s$

Explanation:

From the question w are told that

The mass of the billiard ball A is =m

The initial speed of the billiard ball A = $v_1$ =1 m/s

The mass of the billiard ball B is = 2 m

The initial speed of the billiard ball B = 0

Let the final speed of the billiard ball A = $v_A$

Let The finial speed of the billiard ball B = $v_B$

According to the law of conservation of Energy

$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

$\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

According to the law of conservation of Momentum

$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

$(1 -2v_B)^2 + 2v_B^2 = 1$

$1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

$6v_B^2 -4v_B +1 =1$

$6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

$6v_B -4 = 0$

$v_B = \frac{4}{6}$

$v_B = 0.67 m/s$

4 0

3 years ago

A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change

notsponge [240]

We can solve the problem by using the first law of thermodynamics, which states that:
$\Delta U = Q-W$
where
$\Delta U$ is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
$\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ$

6 0

3 years ago

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

<span>The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>

3 0

2 years ago

If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?

Lady_Fox [76]

·The acceleration of gravity is proportional to

   1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

   = 490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

   (9.8 m/s²) · (1/5)² = 0.39 m/s² .

The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)

= 19.6 newtons .

Just as we expected, his weight at that distance is

   (19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.

7 0

2 years ago