For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3
Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3
Then, the answer is the option C.
Answer:
Explanation:
Given that,
Number of turn N = 40
Diameter of the coil d= 11cm = 0.11m
Then, radius = d/2 = 0.11/2 =0.055m
r = 0.055m
Then, the area is given as
A =πr²
A = π × 0.055²
A = 9.503 × 10^-3 m²
Magnetic Field B = 0.35T
Magnetic field reduce to zero in 0.1s, t = 0.1s
so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).
E.M.F is given as
ε = —N • dΦ/dt
Where magnetic flux is given as
Φ = BA
Then, ε = —N • dΦ/dt
ε = —N • dBA/dt
ε = —NBA/t
Then, its magnitude is
ε = NBA/t
Inserting the values of N, B, A and t
ε = 40×0.35×9.503×10^-3/0.1
ε = 1.33 V
Then, using the relationship between Electric field and electric potential
V = Ed
ε = E•d
E = ε/d
E = 1.33/0.11
E = 12.09 V/m
B! Conduction is touch, so the heat traveled through touch from th stove to the ice cube, therefore melting it
