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Answer:
The rate at which power is generated in the coil is 10.24 Watts
Explanation:
Given;
number of turns of the coil, N = 160
area of the coil, A = 0.2 m²
magnitude of the magnetic field, B = 0.4 T
time for field change = 2 s
resistance of the coil, R = 16 Ω
The induced emf in the coil is calculated as;
emf = dΦ/dt
where;
Φ is magnetic flux = BA
emf = N (BA/dt)
emf = 160 (0.4T x 0.2 m²)/dt
emf = 12.8 V/s
The rate power is generated in the coil is calculated as;
P = V²/ R
P = (12.8²) / 16
P = 10.24 Watts
Therefore, the rate at which power is generated in the coil is 10.24 Watts
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
Resistance = voltage / current.
That's. 120v / 14A = 8.57 ohms.
By the way, voltage doesn't "run through" anything. Current does. That would be the 14 Amps.
