Answer:
Inductance as calculated is 13.12 mH
Solution:
As per the question:
Length of the coil, l = 12 cm = 0.12 m
Diameter, d = 1.7 cm = 0.017 m
No. of turns, N = 235
Now,
Area of cross-section of the wire, A = ![\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20d%5E%7B2%7D%7D%7B4%7D%20%3D%20%5Cfrac%7B%5Cpi%20%5Ctimes%200.017%5E%7B2%7D%7D%7B4%7D%20%3D%202.269%5Ctimes%2010%5E%7B-%204%7D%5C%20m%5E%7B2%7D)
We know that the inductance of the coil is given by the formula:
![L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7Bmu_%7Bo%7DAN%5E%7B2%7D%7D%7Bl%7D%20%3D%20%5Cfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-%207%7D%5Ctimes%202.269%5Ctimes%2010%5E%7B-%204%7D%5Ctimes%20235%5E%7B2%7D%7D%7B0.12%7D%20%3D%201.312%5Ctimes%2010%5E%7B-%204%7D%5C%20H%20%3D%2013.12%5C%20mH)
Answer:
C) 26.6
Explanation:
I don't know how to calculate vector
It'll last 20 hours. If it travels 110 miles in one hours, 110 times 20 equals 2,200.
Answer:
The magnitude of the electric flux is ![3.53\ N-m^2/C](https://tex.z-dn.net/?f=3.53%5C%20N-m%5E2%2FC)
Explanation:
Given that,
Electric field = 2.35 V/m
Angle = 25.0°
Area ![A= 1.65 m^2](https://tex.z-dn.net/?f=A%3D%201.65%20m%5E2)
We need to calculate the flux
Using formula of the magnetic flux
![\phi=E\cdot A](https://tex.z-dn.net/?f=%5Cphi%3DE%5Ccdot%20A)
![\phi = EA\cos\theta](https://tex.z-dn.net/?f=%5Cphi%20%3D%20EA%5Ccos%5Ctheta)
Where,
A = area
E = electric field
Put the value into the formula
![\phi=2.35\times1.65\times\cos 25^{\circ}](https://tex.z-dn.net/?f=%5Cphi%3D2.35%5Ctimes1.65%5Ctimes%5Ccos%2025%5E%7B%5Ccirc%7D)
![\phi=2.35\times1.65\times0.91](https://tex.z-dn.net/?f=%5Cphi%3D2.35%5Ctimes1.65%5Ctimes0.91)
![\phi=3.53\ N-m^2/C](https://tex.z-dn.net/?f=%5Cphi%3D3.53%5C%20N-m%5E2%2FC)
Hence, The magnitude of the electric flux is ![3.53\ N-m^2/C](https://tex.z-dn.net/?f=3.53%5C%20N-m%5E2%2FC)
Answer:
x=0.154kg
Explanation:
(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0
(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0