Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
<h3>
Answer:</h3>
172.92 °C
<h3>
Explanation:</h3>
Concept being tested: Quantity of heat
We are given;
- Specific heat capacity of copper as 0.09 cal/g°C
- Quantity of heat is 8373 calories
- Mass of copper sample as 538.0 g
We are required to calculate the change in temperature.
- In this case we need to know that the amount of heat absorbed or gained by a substance is given by the product of mass, specific heat capacity and change in temperature.
Therefore, to calculate the change in temperature, ΔT we rearrange the formula;
ΔT = Q ÷ mc
Thus;
ΔT = 8373 cal ÷ (538 g × 0.09 cal/g°C)
= 172.92 °C
Therefore, the change in temperature will be 172.92 °C
<span>Vertical lines are 50º apart.
Horizontal lines are 30 minutes apart.</span>
