Answer:
i. 0.2 N ii. 30°
Explanation:
(i) Calculate the magnitude and direction of force on X, when a current of 4A is passed through it.
The magnetic force F = BILsinФ where B = magnetic field strength = 0.1 T, I = current = 4 A and L= length of conductor = 0.5 m. Since the conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic field of 0.1T which is pointing towards the positive Y-axis, both B and L are perpendicular to each other. So, Ф = 90°
So, F = BILsinФ
F = 0.1 T × 4 A ×0.5 m × sin90°
F = 0.1 T × 4 A ×0.5 m × 1
F = 0.2 N
(ii) Through what angle must X be turned in a vertical plane so that the force on X is halved
If F' = BILsinФ' where Ф'=the new angle, and BIL = F
F'/F = sinФ'
Since F'/F = 1/2
sinФ' = 1/2
Ф' = sin⁻¹(1/2) = 30°
The change in internal energy of the gas is

.
In fact, the 1st law of thermodynamics states that the change in internal energy of a system is equal to the amount of heat given to the system (Q) plus the work done on the system (W):

In this example, no work is done on the bottle so W=0, while the heat given to the system is

, so the change in internal energy of the gas is
When the ruler is broughı near the inetal knob, it repels electrons in the metal. Electrons move away froni the ruler and down the metal rod. The knob now has a positive charge. The thin pieces of metal foil at the bottom of the metal rod now have a negative charge.
Answer:
Explanation:
Resistance of the tungsten wire
R = resistivity x length / cross sectional area
= 
= 107 x 10⁻⁴ ohm
Resistance at 120 degree can be obtained from the following formula


= 155.15 x 10⁻⁴ ohm
= 160 x 10⁻⁴ ohm ( rounding off to two syg fig )
current = 12.5
potential diff = 12.5 x 155.15 x 10⁻⁴ V
= 0 .1939 V
= .19 V
required electric field = potential diff / length of wire
= .1939 / 16 x 10⁻²
= 1.2 N / C
D) It isn't accelerating
This is because if both sides are pushing with the exact same force, than one force is not overpowering the other, therefor the wheelbarrow would not be moving.