Answer:
Inductance, L = 0.0212 Henries
Explanation:
It is given that,
Number of turns, N = 17
Current through the coil, I = 4 A
The total flux enclosed by the one turn of the coil, ![\phi=5\times 10^{-3}\ Tm^2](https://tex.z-dn.net/?f=%5Cphi%3D5%5Ctimes%2010%5E%7B-3%7D%5C%20Tm%5E2)
The relation between the self inductance and the magnetic flux is given by :
![L=\dfrac{N\phi}{I}](https://tex.z-dn.net/?f=L%3D%5Cdfrac%7BN%5Cphi%7D%7BI%7D)
![L=\dfrac{17\times 5\times 10^{-3}}{4}](https://tex.z-dn.net/?f=L%3D%5Cdfrac%7B17%5Ctimes%205%5Ctimes%2010%5E%7B-3%7D%7D%7B4%7D)
L = 0.0212 Henries
So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.
The correct answer is B. Nitrogen
Answer:
a
![l = 0.305 \ m](https://tex.z-dn.net/?f=l%20%3D%200.305%20%5C%20%20m)
b
![f = 3.0*10^{11} \ Hz](https://tex.z-dn.net/?f=f%20%3D%203.0%2A10%5E%7B11%7D%20%5C%20%20Hz)
Explanation:
From the question we are told that
The wavelength is ![\lambda = 12.2 \ cm = 0.122 \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%2012.2%20%5C%20%20cm%20%20%3D%200.122%20%5C%20%20m)
The number of antinodal planes of the electric field considered is n = 5
The width is mathematically represented as
![l = \frac{ n \lambda}{2}](https://tex.z-dn.net/?f=l%20%20%3D%20%20%5Cfrac%7B%20n%20%5Clambda%7D%7B2%7D)
![l = \frac{5 * 0.122 }{ 2}](https://tex.z-dn.net/?f=l%20%3D%20%5Cfrac%7B5%20%2A%200.122%20%7D%7B%202%7D)
![l = 0.305 \ m](https://tex.z-dn.net/?f=l%20%3D%200.305%20%5C%20%20m)
Generally the frequency the errors was made is mathematically represented as
![f = \frac{c}{\lamda_k}](https://tex.z-dn.net/?f=f%20%3D%20%20%5Cfrac%7Bc%7D%7B%5Clamda_k%7D)
Here c is the speed of light with value ![c = 3.0*10^{8} \ m/s](https://tex.z-dn.net/?f=c%20%3D%20%203.0%2A10%5E%7B8%7D%20%5C%20%20m%2Fs)
is the wavelength of the microwave has to be in order for there still to be five antinodal planes of the electric field along the width of the oven, which is mathematically represented as
![\lambda_k = \frac{ \lambda * \frac{0.04}{2} }{n/2}](https://tex.z-dn.net/?f=%5Clambda_k%20%20%3D%20%20%5Cfrac%7B%20%5Clambda%20%2A%20%20%5Cfrac%7B0.04%7D%7B2%7D%20%7D%7Bn%2F2%7D)
![\lambda_k = \frac{0.122*0.02}{5/2}](https://tex.z-dn.net/?f=%5Clambda_k%20%20%3D%20%20%5Cfrac%7B0.122%2A0.02%7D%7B5%2F2%7D)
So
![f = \frac{3.0*10^{8}}{0.000976}](https://tex.z-dn.net/?f=f%20%3D%20%20%5Cfrac%7B3.0%2A10%5E%7B8%7D%7D%7B0.000976%7D)
![f = 3.0*10^{11} \ Hz](https://tex.z-dn.net/?f=f%20%3D%203.0%2A10%5E%7B11%7D%20%5C%20%20Hz)
Answer:
21000 N
Explanation:
From the question given above, the following data were obtained:
Change in momentum = 105000 kg.m/s
Time = 5s
Force =?
Force is related to momentum and time according to the following formula:
Force = Change in momentum / time
With the above formula, we can calculate the force the white car experience during the collision. This can be obtained as illustrated below:
Change in momentum = 105000 kg.m/s
Time = 5s
Force =?
Force = Change in momentum / time
Force = 105000 / 5
Force = 21000 N
Thus, the white car experience a force of 21000 N during the collision.
The car has an average acceleration of
![a_{\rm ave} = \dfrac{\Delta v}{\Delta t} = \dfrac{0\frac{\rm m}{\rm s} - 20\frac{\rm m}{\rm s}}{4\,\mathrm s} = -5\dfrac{\rm m}{\mathrm s^2}](https://tex.z-dn.net/?f=a_%7B%5Crm%20ave%7D%20%3D%20%5Cdfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%20%3D%20%5Cdfrac%7B0%5Cfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%20-%2020%5Cfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%7D%7B4%5C%2C%5Cmathrm%20s%7D%20%3D%20-5%5Cdfrac%7B%5Crm%20%20m%7D%7B%5Cmathrm%20s%5E2%7D)
Then, by Newton's second law, the average force exerted by the brakes on the car is
![F_{\rm ave} = ma_{\rm ave} = (1200\,\mathrm{kg})\left(-5\dfrac{\rm m}{\mathrm s^2}\right) = \boxed{-6000 N}](https://tex.z-dn.net/?f=F_%7B%5Crm%20ave%7D%20%3D%20ma_%7B%5Crm%20ave%7D%20%3D%20%281200%5C%2C%5Cmathrm%7Bkg%7D%29%5Cleft%28-5%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%20%3D%20%5Cboxed%7B-6000%20N%7D)
which is to say, the average force has a magnitude of 6000 N and is directed opposite the car's motion.