Answer:
a) 0.09N b) positive x direction
Explanation:
Force on a conductor carrying current in a magnetic field can be expressed as;
F = BILsin(theta) where
F is the force on the conductor (wire)
B is the uniform magnetic field I'm Tesla = 1.8Tesla
I is the current in the wire = 5×10^-2A
L is the length of the wire = 1m
theta is the angle that the conductor make with the magnetic field = 90° (since the wire in the horizontal direction is perpendicular to the field acting upwards)
Substituting this value in the formula to get F we have;
F = 1.8×5×10^-2×1 × sin90°
F = 0.09N
The force on the wire is 0.09N
b) The direction of the force is in the positive x direction since the wire acts horizontally to the magnetic field.
Answer:

Explanation:
A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a
In this way we obtain that:

In this way,


In this way the entropy change is,

B) Whenever entropy yields a positive result, the process can be carried out adiabatically.
Inside the torch, they usually use copper or brass to conduct electricity.
<h2>
Option C is the correct answer.</h2>
Explanation:
Gravitational force is given by

G=6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²
M = Mass of object 1
m = mass of object 2
r = Distance between objects.
Here only variable is r value.
In case 1

In case 2

Option C is the correct answer.
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5