Answer:
So you can get the results correctly
Answer:
The distance is ![D = 2.6 \ m](https://tex.z-dn.net/?f=D%20%20%3D%20%202.6%20%5C%20m)
Explanation:
From the question we are told that
The wavelength of the light is ![\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20476.1%20%5C%20nm%20%20%3D%20%20476.1%20%2A10%5E%7B-9%7D%20%5C%20m)
The distance between the slit is ![d = 0.29 \ mm = 0.29 *10^{-3} \ m](https://tex.z-dn.net/?f=d%20%3D%20%200.29%20%5C%20%20mm%20%20%3D%20%200.29%20%2A10%5E%7B-3%7D%20%5C%20m)
The between the first and second dark fringes is ![y = 4.2 \ mm = 4.2 *10^{-3} \ m](https://tex.z-dn.net/?f=y%20%3D%20%204.2%20%5C%20mm%20%20%3D%20%204.2%20%2A10%5E%7B-3%7D%20%5C%20m)
Generally fringe width is mathematically represented as
![y = \frac{\lambda * D }{d}](https://tex.z-dn.net/?f=y%20%20%3D%20%20%5Cfrac%7B%5Clambda%20%2A%20D%20%7D%7Bd%7D)
Where D is the distance of the slit to the screen
Hence
![D = \frac{y * d}{\lambda }](https://tex.z-dn.net/?f=D%20%20%3D%20%20%5Cfrac%7By%20%2A%20%20d%7D%7B%5Clambda%20%7D)
substituting values
![D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }](https://tex.z-dn.net/?f=D%20%20%3D%20%20%5Cfrac%7B%204.2%20%2A10%5E%7B-3%7D%20%2A%20%20%200.29%20%2A10%5E%7B-3%7D%7D%7B%20476.1%20%2A10%5E%7B-9%7D%20%7D)
![D = 2.6 \ m](https://tex.z-dn.net/?f=D%20%20%3D%20%202.6%20%5C%20m)
Answer:
the hypotenuse = 13.78 cm
Ф = 27.44°
θ = 62.56°
explanation:
Answer:
Np=130 turns, Is=220 A
Explanation:
Part A:
Ns=100 turns
Vp= 1.5*104 V = 156 V
Vs= 120 V
As Np/Ns= Vp/Vs
⇒ Np= ![\frac{Vp}{Vs}*Ns](https://tex.z-dn.net/?f=%5Cfrac%7BVp%7D%7BVs%7D%2ANs)
⇒Np=![\frac{156*100}{120}](https://tex.z-dn.net/?f=%5Cfrac%7B156%2A100%7D%7B120%7D)
Np= 130 turns (It has two significant figures)
Part B
Is=280 A, Vs=120 V, Vp= 1.5*104 V= 156 V
For ideal transformer
Pin=Pout
Ip*Vp= Is*Vs
⇒Ip=(
)
⇒Ip=![\frac{280*120}{156}](https://tex.z-dn.net/?f=%5Cfrac%7B280%2A120%7D%7B156%7D)
⇒Ip=215.4 A
to represent it as two significant figures we will write as
⇒Ip=220 A