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Viktor [21]
4 years ago
13

The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A:

Physics
1 answer:
Phantasy [73]4 years ago
5 0

Answer:

I believe it will it is B and D. Because they have the same shape and on both the cart wil be speeding up.

I HOPE THIS HELPS! HAVE A GREAT DAY! :)

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We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goe
o-na [289]

Answer:

See answer

Explanation:

Given quantities:

\eta = 0.05\\ W=90[W]\\r=0.0285[m]

where \eta is the efficiency of the lightbulb (visible light is 5% of the total power), W is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

I =\frac{P}{A}

a) Now the effective power is \eta*W, therefore:

I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]

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]

now E and B are related:

E=cB\\ B=\frac{E}{c} and c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}

replace in S_{average}

S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]

we replace the values and we get:

E= \sqrt{\frac{2I}{\epsilon_{0}c}}

E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]

therefore

B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]

8 0
4 years ago
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