Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ
Answer:
(a) 42.28°
(b) 37.08°
Explanation:
From the principle of refraction of light, when light wave travels from one medium to another medium, we have:
= sinθ/sinθ
In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.
For critical angle θ = θ, θ = 90°;
(a)
= sinθ/sin90°
0.6728 = sinθ
= sinθ/sin90°
0.60299 = sinθ[tex]_{c}
θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°
Answer:
0.182 m/s
Explanation:
m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s
let the velocity of loaded freight car is v
Use the conservation of momentum
m1 x u1 + m2 x 0 = (m1 + m2) x v
30,000 x 0.85 = (30,000 + 110,000) x v
v = 0.182 m/s
Explanation:
w = (4.52 ± 0.02) cm, x = ( 2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x + y - w and its uncertainty.
z = x + y - w = 2.0 + 3.0 - 4.5 = 0.5 cm
Dz = Dx + Dy + Dw = 0.2 + 0.6 + 0.02 = 0.82 rounding to 0.8 cm
So z = (0.5 ± 0.8) cm
Solution with standard deviations, Eq. 1b, Dz = 0.633 cm
z = (0.5 ± 0.6) cm
Notice that we round the uncertainty to one significant figure and round the answer to match.