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Brut [27]
2 years ago
9

PLEASE ASAP ILL GIVE BRAINLIEST.

Physics
1 answer:
boyakko [2]2 years ago
4 0
I think it’s number 3
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When energy is added to a wave, how can the wave change?.
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When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.

Explanation:

When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.

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A car accelerates at a rate of 4 m/s in 5 s. Calculate the change of velocity PLS HELP
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Answer:

Explanation:

a=Δv/Δt

Δv=a*Δt

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A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

9 0
3 years ago
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