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2 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]2 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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A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

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(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
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<span>0 = 20 + a(4)

</span><span>4 x a = -20
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so, the answer is <span>-5 m/s^2. or -5 meter per second</span>

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four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?

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There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.

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In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being

german

Answer: Got It!

<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.

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2 years ago

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

4 0

2 years ago

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

<u>The relation between the change in pressure with height is given as:</u>

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

<em>Now putting the respective values:</em>

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0

1 year ago