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2 years ago

30pts!!!06.01 Simple Harmonic Motion Lab: Virtual Lab Simulation

Physics

2 answers:

Ostrovityanka [42]2 years ago

8 0

Answer:

Hello; I would like to say you that brainly is for short answers not very long answers if you want the answer you can contact brainly services

mel-nik [20]2 years ago

8 0

Answer:

https://prezi.com/020q6srszjtp/simple-harmonic-motion-lab/

Explanation:

i found a prezi slide show for this online with answers and explanations :)

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Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

If the moon phase is seen as a waxing crescent moon in london, what phase of the moon would be seen in new york if it's viewed a

Murljashka [212]

Lunar phase is the same wherever on Earth you observe
<span>Last (third) quarter rises at midnight, sets at noon. </span>
<span>First quarter rises at noon, sets at midnight</span>

4 0

2 years ago

What is the purpose of humping

stiv31 [10]

Explanation:

I'm not sure to be honest lol

4 0

2 years ago

Read 2 more answers

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Arte-miy333 [17]

Answer:

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Most asteroids revolve around the sun in an orbit between those of Mars and Jupiter.

They form a wide band called the Asteroid belt.

Other asteroids have orbits that cross Earth’s orbit. These asteroids are called Earth-crossers.

Asteroids probably consist of matter that never agglomerated into a planet when the solar system was forming.

The comet’s core is composed of ice and dust.

Comets heat up and begin to change from a solid to a gas as they approach the sun.

The matter surrounding a comet’s core is vaporized and forms a very bright halo of ice or dust not sure, and an enormous cloud of dust or gasses not sure envelopes the head of the comet.

7 0

3 years ago

¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una

ValentinkaMS [17]

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

5 0

3 years ago