Answer:
Second order line appears at 43.33° Bragg angle.
Explanation:
When there is a scattering of x- rays from the crystal lattice and interference occurs, this is known as Bragg's law.
The Bragg's diffraction equation is :
.....(1)
Here n is order of constructive interference, λ is wavelength of x-ray beam, d is the inter spacing distance of lattice and θ is the Bragg's angle or scattering angle.
Given :
Wavelength, λ = 1.4 x 10⁻¹⁰ m
Bragg's angle, θ = 20°
Order of constructive interference, n =1
Substitute these value in equation (1).
d = 2.04 x 10⁻¹⁰ m
For second order constructive interference, let the Bragg's angle be θ₁.
Substitute 2 for n, 2.04 x 10⁻¹⁰ m for d and 1.4 x 10⁻¹⁰ m for λ in equation (1).
<em>θ₁ </em>= 43.33°
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
That's a weird graph, but judging from the units the acceleration is the slope of the graph.
a = (0.8 - 0.3)/(0.16 - 0.055) = 4.76 m/s²
Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.
W is the work done by the friction.
v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.
