Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
Answer:
1.08
Explanation:
This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then
path difference created will be 2μ t.
For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference
path diff = nλ , for minimum t , n =1
path diff = λ
2μ t. = λ
μ = λ / 2t
= 626 / 2 x 290
= 1.08
Weight = (mass) x (gravity)
= (2.2 kg) x (3.7 m/s^2)
= 8.1 newtons. (about 1.84 pounds)
1 kg of water = 1 L = 1 dm³
762 g of water = 762 cm³
Current= V/R
C=60 volts/12 ohms
C= 5 amperes
Hope I helped :)
