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3 years ago

10

In the deep ocean, a water wave with wavelength 95 m travels at 12 m/s. Suppose a small boat is at the crest of this wave, 1.2 m

above the equilibrium position. What will be the vertical position of the boat 5.0 s later?

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

the vertical position is 1.1971m

Explanation:

Recall that

$y = 1.2 * cos (\frac{2*\pi * distance travelled}{wavelenght})$

recall that $v = \frac{distance}{time}$

$thus; distance = v* t$

this implies that distance = 12 * 5.0

= 60

therefore; $y = 1.2 cos \frac{2*3.142*60}{95}$

y = 1.1971m

the

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2 years ago

A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and,

Doss [256]

Answer:

-0.00152 V

Explanation:

Parameters given:

Diameter of the loop = 11 cm = 0.11m

Rate of change of magnetic field, dB/dt = 0.16 T/s

Radius of the loop = 0.055m

The area of the loop will be:

A = pi * r²

A = 3.142 * 0.055²

A = 0.0095 m²

The EMF induced in a loop of wire due to the presence of a changing magnetic field, dB, in a time interval, dt, is given as:

EMF = - N * A * dB/dt

In this case, there's only one loop, so N = 1.

Therefore:

EMF = -1 * 0.0095 * 0.16

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3 years ago

Read 2 more answers

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

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Answer:

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