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3 years ago

7

The type of circuit in which there is only one path for an electric current to flow is known as a _____ circuit. Question 6 opti

ons: Parallel Series Light Component

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

series circuit

Explanation:

A series circuit is wired with only one path for the current to flow through all the devices in a row and back to the starting point.

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A current of 0.96 amps flows through a copper wire 0.44 mm in diameter when connected to a potential difference of 15 v. how lon

FinnZ [79.3K]

By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
$V=IR$
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
$R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega$

Now we can use the following equation to calculate the length of the wire:
$R= \frac{\rho L}{A}$ (1)
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity $\rho=1.68 \cdot 10^{-8} \Omega m$ . The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m$
And the cross-sectional area is
$A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2$

So now we can rearrange eq.(1) to calculate the length of the wire:
$L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m$

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aalyn [17]

2. Newton's laws of motion

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3 years ago

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Plz help it’s passed due!

pickupchik [31]

It is b 9.1 :) have a nice day

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4 years ago

A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi

iVinArrow [24]

Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

$F = \dfrac{GMm}{r^2}$ when r>R

When Point is inside the Sphere

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$ when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

$F = \dfrac{GMm}{r^2}$ [/tex]

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}$

F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}$

$F = 5.68\times 10^{-9}\ N$

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3 years ago

A large group of people get together. each one rolls a die 180 times, and counts the number of 1's. about what percentage of the

Arte-miy333 [17]

<span>Around 99% - 100%. This is because a die is six sided, which gives the odds of a one coming up being roughly 17% independent of every roll. 17% of 180 trials comes out to 30-31 times a one will show up every 180 trials. This puts you right in the middle of the 15-45 range which means that somebody will almost ALWAYS reach 15-45 one's in a trial of 180 rolls.</span>

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3 years ago