Answer:
Explanation:
E₀ = 229.1 V/m
E = E₀ / √2 = 229.1 / 1.414 = 162 V/m
B = E / c ( c is velocity of em waves )
= 162 / (3 x 10⁸) = 54 x 10⁻⁸ T
rate of energy flow = ( E x B ) / μ₀
= 162 x 54 x 10⁻⁸ / 4π x 10⁻⁷
= 69.65 W per m².
Lubricants, Magnetic Levitation, and Ball Bearing.
Answer:
V = 493421.05 [gal]
Explanation:
This is a problem that consists of handling units, we can calculate by first-hand the volume, then convert units from cubic meters to gallons.
V = 50 * 25 * 1.5
V = 1875 [m^3]
Now we need to convert units, using the proper conversion factor.
![1875[m^3]*\frac{1000lt}{1m^3} *\frac{1gal}{3.8lt} \\493421.05[gal]](https://tex.z-dn.net/?f=1875%5Bm%5E3%5D%2A%5Cfrac%7B1000lt%7D%7B1m%5E3%7D%20%2A%5Cfrac%7B1gal%7D%7B3.8lt%7D%20%5C%5C493421.05%5Bgal%5D)
solution:
the kinetic energy acquired by the electron when it is accelerated y an electric field is
k=5.25\times10^-16j
the kinetic energy acquired by the electron will ne proportional to the potential difference between the plates across which the electric field is applied.it is given by,
e\Delta v = k
here,\delta v is the potential difference between the plates, e is the charge on an electron and k is the kinetic energy.
reassange the above expression
\delta v = \frac{k}{e}
the potential difference between the plates will be
\delta v = \frac{k}{e}
sunstitute 5.25 \times 10^-16j for k and 1.6 \times10^-19c for e is the above equation
\delta v =\frac{5.25\times10^-16j}{1.60\times10^-19c}
=3.28\times10^3v(\frac{1kv}{1000v})
=3.28kv
since the electrons will e accelerated towards the plate at higher potential.
hance p;ate b will be at higher potential
