Answer:
See answer
Explanation:
Given quantities:
![\eta = 0.05\\ W=90[W]\\r=0.0285[m]](https://tex.z-dn.net/?f=%5Ceta%20%3D%200.05%5C%5C%20W%3D90%5BW%5D%5C%5Cr%3D0.0285%5Bm%5D)
where 
is the efficiency of the lightbulb (visible light is 5% of the total power), 
is the total power of the lightbulb, r is the radius of the lightbulb in meters.
Intensity is power divided by area:
a) Now the effective power is 
, therefore:
![I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]](https://tex.z-dn.net/?f=I%20%3D%5Cfrac%7B%5Ceta%2AW%7D%7B%5Cpi%20r%5E2%7D%3D%5Cfrac%7B0.05%2A90%7D%7B4%5Cpi%20%280.0285%29%5E2%7D%3D440.87%5BW%2Fm%5E2%5D)
b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:
![S_{average}= \frac{EB}{2\mu_{0}}[W/m]](https://tex.z-dn.net/?f=S_%7Baverage%7D%3D%20%5Cfrac%7BEB%7D%7B2%5Cmu_%7B0%7D%7D%5BW%2Fm%5D)
now 
and 
are related:
and 
replace in 
![S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]](https://tex.z-dn.net/?f=S_%7Baverage%7D%3DI%3D%20%5Cfrac%7Bc%20%5Cepsilon_%7B0%7DE%5E2%7D%7B2%7D%5BW%2Fm%5D)
we replace the values and we get:
![E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]](https://tex.z-dn.net/?f=E%20%3D%20%5Csqrt%7B%5Cfrac%7B2%28440.8%29%7D%7B8.85%2A10%5E%7B-12%7D3%2A10%5E8%7D%7D%3D576.24%5BV%2Fm%5D)
therefore
![B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]](https://tex.z-dn.net/?f=B%3D%5Cfrac%7BE%7D%7Bc%7D%3D%5Cfrac%7B576.24%7D%7B3%2A10%5E%7B8%7D%7D%3D1.92%2A10%5E%7B-6%7D%5BT%5D)
Sometimes if you have the word per, you need to multiply. In this problem, they are asking you to find the amount of meters a snail can travel per minute.They are giving you the amount of minutes, so in this case you multiply 15 times 0.30. The answer is 4.5 meters.
Answer:
Definitely without a doubt a pumpkin
Answer:
I think the answer is half a second cause Time period = Time taken by the pendulum/Number of oscillations made by the pendulum so 5 divided by 10 equals 0.5 which is half a second
