Answer:
1.7135 x 10^10 J / s
Explanation:
R = 20 cm = 0.2 m
e = 0.79
To = 15 degree C = 15 + 273 = 288 K
T = 200 degree C = 200 + 273 = 473 K
Energy radiated per unit time = A x e x (T^4 - To^4)
Power radiated = 4 x 3.14 x 0.2 x 0.2 x 0.79 x (473^4 - 288^4)
= 1.7135 x 10^10 J / s
Answer:
KE = 1/2mv^2
KE = 1/2(1120)(40^2)
KE = 560(1600)
KE = 896000
Let me know if this helps!
ΔVl = L di/dt
i = i₀e -t/T
di/dt = i₀ × (-1/T) e -t/T
ΔVl = L× (-I/T i₀e -t/T
ΔVl = -L/T i₀e -t/T
b. 15mm, i₀ = 36mA, T = 1.1m
t= Os
ΔVl = 0,491V
C. t = 1ms
ΔVl = 0.198V
t = 2ms
ΔVl = 0.08V
E. t = ms
ΔVl = 0.032V
Answer:
Explanation:
Let us assume that the radius of the wire, r = 0.8 mm = 0.0008 m
EMF of the battery, V = 12 V
Slope of I versus 1/d, m = 600 A-m
The resistance of any material is given by :
d is the length of wire
Since,
So, the resistivity of the material of which the wire is made is .