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3 years ago

Define and describe the following terms: amplitude, wavelength, frequency, period, tension.

Physics

1 answer:

maw [93]3 years ago

7 0

Answer:

Amplitude—distance between the resting position and the maximum displacement of the wave

Frequency—number of waves passing by a specific point per second

Period—time it takes for one wave cycle to complete

wavelength λ - the distance between adjacent identical parts of a wave, parallel to the direction of propagation.

Tension - described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar one-dimensional continuous object, or by each end of a rod, truss member, or similar three-dimensional object

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svlad2 [7]

Answer:

Explanation:

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2 years ago

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3 years ago

An elastic conducting material is stretched into a circular loop of 9.65 cm radius. It is placed with its plane perpendicular to

Nadya [2.5K]

Answer:

The induced emf in the coil is 0.522 volts.

Explanation:

Given that,

Radius of the circular loop, r = 9.65 cm

It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s , $\dfrac{dr}{dt}=-0.756\ m/s$

Due to the shrinking of radius of the loop, an emf induced in it. It is given by :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=B\dfrac{-d(\pi r^2)}{dt}\\\\\epsilon=2\pi rB\dfrac{dr}{dt}\\\\\epsilon=2\pi \times 9.65\times 10^{-2}\times 1.14\times 0.756\\\\\epsilon=0.522\ V$

So, the induced emf in the coil is 0.522 volts.

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3 years ago

A lead nucleus is spherical with a radius of about 7 ✕ 10⁻¹⁵ m. The nucleus contains 82 protons (and typically 126 neutrons). Be

Komok [63]

Answer:

∈= $3.1584x10^{26} \frac{V}{m}$

Explanation:

Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus

∈ $=\frac{P}{E_{o}}$

The P is the charge density and 'Eo' is the constant of permittivity in free space

to find P

$P=\frac{q}{V}$

$V=\frac{4}{3}*\pi*r^3$

$V=\frac{4}{3}\pi*(7x10^{-15})^3$

$V=2.932x10^{-14} m^3$

$P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}$

So replacing

∈ $=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}$

∈= $3.1584x10^{26} \frac{V}{m}$

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