Answer:
(a) A = 
(b) 
(c) 
(d) 
Solution:
As per the question:
Radius of atom, r = 1.95
Now,
(a) For a simple cubic lattice, lattice constant A:
A = 2r
A = 
(b) For body centered cubic lattice:


(c) For face centered cubic lattice:


(d) For diamond lattice:


<span>If two waves with identical crests and troughs meet, what is happening?
</span>C. Constructive interference is occurring.
Answer:
the direction of acceleration of the vehicle is the same direction of its velocity of car
s acceleration has the opposite direction to the car speed.
Explanation:
The initial acceleration of the car can be calculated with
v = v₀ + a t
a = (v-v₀) t
indicate that the initial velocity is zero (v₀ = 0 m / s)
a = v / t
a = 300 / t
the direction of acceleration of the vehicle is the same direction of its acceleration movement.
When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small
Answer:
0.01154 A
Explanation:
We have given the energy in the magnetic field
Value of inductance L =0.060 H
Energy stored in magnetic field is given by 


So the current flowing through rectangular toroid will be 0.01154 A
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant