Hi there!
p(a∩b) = 0.
Since a and b are mutually exclusive events, there is no overlap or intersection of events.
These two events CANNOT both occur, so the intersection of p(a) and p(b) does not exist.
Thus, p(a∩b) = 0.
Complete Question
The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air
Answer:
The value is ![P = 3.270960 *10^{6} \ W](https://tex.z-dn.net/?f=P%20%3D%203.270960%20%2A10%5E%7B6%7D%20%5C%20%20W)
Explanation:
From the question we are told that
The electric field strength is ![E = 4.0 \ M \ V/m = 4.0 *10^6 \ V/m](https://tex.z-dn.net/?f=E%20%20%3D%204.0%20%5C%20%20M%20%5C%20%20V%2Fm%20%20%3D%20%204.0%20%2A10%5E6%20%5C%20%20V%2Fm)
The diameter is ![d = 1.4 \ cm = 0.014 \ m](https://tex.z-dn.net/?f=d%20%3D%20%201.4%20%5C%20%20cm%20%20%3D%200.014%20%5C%20m)
Generally the radius is mathematically represented as
![r = \frac{d}{2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Bd%7D%7B2%7D)
=> ![r = \frac{0.014}{2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B0.014%7D%7B2%7D)
=> ![r = 0.007 \ m](https://tex.z-dn.net/?f=r%20%20%20%3D%200.007%20%5C%20%20m)
Generally the cross-sectional area is mathematically represented as
![A = \pi r^2](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cpi%20r%5E2)
![A = 3.142 * (0.007)^2](https://tex.z-dn.net/?f=A%20%3D%20%203.142%20%2A%20%20%280.007%29%5E2)
![A = 0.000154 \ m^2](https://tex.z-dn.net/?f=A%20%3D%200.000154%20%5C%20m%5E2)
Generally the maximum power that can be delivered is mathematically represented as
![P = \frac{c * \epsilon_o * E^2 * A }{2}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7Bc%20%2A%20%20%5Cepsilon_o%20%20%2A%20%20E%5E2%20%2A%20%20A%20%7D%7B2%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 permittivity of free space with value ![\epsilon_o = 8.85 *10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2](https://tex.z-dn.net/?f=%5Cepsilon_o%20%20%3D%20%208.85%20%2A10%5E%7B-12%7D%20%20%5C%20m%5E%7B-3%7D%20%5Ccdot%20kg%5E%7B-1%7D%5Ccdot%20%20s%5E4%20%5Ccdot%20A%5E2)
![P = \frac{3,0*10^8 * 8.85*10^{-12} * (4 *10^6)^2 * 0.00154}{ 2}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B3%2C0%2A10%5E8%20%2A%20%208.85%2A10%5E%7B-12%7D%20%2A%20%20%284%20%2A10%5E6%29%5E2%20%2A%200.00154%7D%7B%202%7D)
![P = 3.270960 *10^{6} \ W](https://tex.z-dn.net/?f=P%20%3D%203.270960%20%2A10%5E%7B6%7D%20%5C%20%20W)
Answer:
Liquid
Explanation: It can't be gas because gases are all around us and we can still see light. For solids it can't even move through a solid. Ex: If you're in a solid dome made of rock you can't get any light through. Let me know if I'm right!
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
Answer:
The current in primary coil is 2.08 A.
Explanation:
Given that,
Power = 500 W
Voltage = 16 kV
Number of turns = 80000
We need to calculate the number of current
Using formula of voltage
![\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B1%7D%7D%7BN_%7B2%7D%7D%3D%5Cdfrac%7BV_%7B1%7D%7D%7BV_%7B2%7D%7D)
Put the value into the formula
![\dfrac{N_{1}}{80000}=\dfrac{120}{16\times10^{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_%7B1%7D%7D%7B80000%7D%3D%5Cdfrac%7B120%7D%7B16%5Ctimes10%5E%7B3%7D%7D)
![N_{1}=\dfrac{240}{16\times10^{3}}\times80000](https://tex.z-dn.net/?f=N_%7B1%7D%3D%5Cdfrac%7B240%7D%7B16%5Ctimes10%5E%7B3%7D%7D%5Ctimes80000)
![N_{1}=1200](https://tex.z-dn.net/?f=N_%7B1%7D%3D1200)
We need to calculate the current in secondary coil
Using formula of current
![i=\dfrac{P}{V_{2}}](https://tex.z-dn.net/?f=i%3D%5Cdfrac%7BP%7D%7BV_%7B2%7D%7D)
![i_{2}=\dfrac{500}{16\times10^{3}}](https://tex.z-dn.net/?f=i_%7B2%7D%3D%5Cdfrac%7B500%7D%7B16%5Ctimes10%5E%7B3%7D%7D)
![i_{2}=0.03125\ A](https://tex.z-dn.net/?f=i_%7B2%7D%3D0.03125%5C%20A)
We need to calculate the current in primary coil
![\dfrac{I_{1}}{I_{2}}=\dfrac{N_{2}}{N_{1}}](https://tex.z-dn.net/?f=%5Cdfrac%7BI_%7B1%7D%7D%7BI_%7B2%7D%7D%3D%5Cdfrac%7BN_%7B2%7D%7D%7BN_%7B1%7D%7D)
Put the value into the formula
![\dfrac{i_{1}}{0.03125}=\dfrac{80000}{1200}](https://tex.z-dn.net/?f=%5Cdfrac%7Bi_%7B1%7D%7D%7B0.03125%7D%3D%5Cdfrac%7B80000%7D%7B1200%7D)
![i_{1}=\dfrac{80000}{1200}\times0.03125](https://tex.z-dn.net/?f=i_%7B1%7D%3D%5Cdfrac%7B80000%7D%7B1200%7D%5Ctimes0.03125)
![i_{1}=2.08\ A](https://tex.z-dn.net/?f=i_%7B1%7D%3D2.08%5C%20A)
Hence, The current in primary coil is 2.08 A.