Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
The answer is 403 ml coz when you divide 0.403 with 1000 ml you get 403 ml
Answer:
Most likely, it will be harder to get strong magnets to change phase because they have more density.
Answer:
Explanation:
Given
mass of First Block 
Temperature 
mass of second block 
Temperature 
Heat capacity of aluminium c=899 J/kg-K
Final Temperature acquired by both blocks at steady state
Heat loss first block =Heat gain by second block
