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3 years ago

7

To start the analysis of this circuit you must write energy conservation (loop) equations. Each equation must involve a round-tr

ip path that begins and ends at the same location. Each segment of the path should go through a wire, a bulb, or a battery (not through the air). How many valid energy conservation (loop) equations is it possible to write for this circuit?

1 answer:

REY [17]3 years ago

6 0

Answer:

Explanation:

Electric field talks about a region around a charged particle or object within which a force would be exerted on other charged particles or objects. to find the electric field inside the bulb we will apply the electric filed formula.

Please kindly check attachment for step by step explaination.

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

A-delta fibers : A) are small, myelinated fibers. B) transmit pain signals at a slower rate than C-fibers. C) typically transmit

daser333 [38]

Answer:

A

Explanation:

Aδ fibers carry cold, pressure, and acute pain signals, and because they are thin (2 to 5 μm in diameter) and myelinated, they send impulses faster than unmyelinated C fibers, but more slowly than other, more thickly myelinated group A nerve fibers. Their conduction velocities are moderate.

7 0

3 years ago

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi

USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

$f=\frac{nv}{2l}\\$ .

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

$f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\$ .

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

$f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz$

Hence the require beat is

$B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz$ .

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

$L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204$

Now to determine the extension,

$L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\$

4 0

3 years ago

You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find

wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0

3 years ago

What do you think that astronomers mean when they use the term observable universe?

iren2701 [21]

The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us.

Hope this helps :)

6 0

3 years ago