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2 years ago

14

How does time relate to motion

1 answer:

docker41 [41]2 years ago

5 0

As we move, time goes up. Think of it on a graph; as time increases on the x axis, motion can either stay the same, increase, or decrease.

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If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is th

Taya2010 [7]

Answer: The magnitude of the current in the second wire 2.67A

Explanation:

Here is the complete question:

Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?

Explanation: Please see the attachments below

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3 years ago

Atoms can join together to form ______ substance

ELEN [110]

Answer:

Molecules

Explanation:

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3 years ago

The amount of light lost within a fiber-optic system is known as

densk [106]

Attenuation is the correct answer.

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3 years ago

If a runner goes 500 meters with an average speed of 7.0 m/s. How long were they running?

ivann1987 [24]

Answer: s=d/t= 500/7.0=71.42 m/s

Explanation:

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3 years ago

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

So

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago