Answer:
C. Just measure the volume of the container it is in
Explanation:
Another why of measuring the volume of gas is by filling a contractor with water then in invert a glass jar air will miss place the space taken by water then measure the volume of water misplaced to get the volume to air
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)
The second law states that the total entropy can never decrese over time for an isolated system
Answer:
2.0 m
Explanation:
Energy is conserved.
Initial KE = Final PE + Work done by friction
½ mv² = mgh + Fd
½ mv² = mgh + mgμd
½ v² = gh + gμd
½ v² − gh = gμd
d = (½ v² − gh) / (gμ)
d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))
d = 2.0 m