PHYSICS COLLEGE

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?

Answers

Answer 1

Answer:

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $(velocity)/(2 *length)$

velocity = $\sqrt{(tension)/(mass per unit length) }$

mass per unit length= $(3.5)/(1000*1.22)$ =0.00427 $(kg)/(m)$

Now calculating velocity v= $\sqrt{(255)/(0.00427) }$

=244.3 $(m)/(sec)$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $(244.3)/(2 *0.7)$ =174.5 hz

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Why do electrical devices have resistance

Answers

As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?

Answers

Answer:

a). $E_(kA)=1.2635 J$

b). $V_(B)=4.535(m)/(s)$

c). Δ $E_(t)=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_(kA)=(1)/(2)*m*v_(A) ^(2) \n v_(A)=1.9 (m)/(s)\n m=0.70kg\nE_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \nE_(kA)=1.2635 J$

b).

$E_(kB)=(1)/(2)*m*v_(B) ^(2)$

$V_(B)^(2)=(E_(kB)*2)/(m) \nV_(B)=\sqrt{(E_(kB)*2)/(m)} \nV_(B)=\sqrt{(7.2J*2)/(0.70kg)} \nV_(B)=4.53 (m)/(s)$

c).

net work= EkA+EkB

$E_(t)=E_(kA)+ E_(kB)\nE_(t)=1.2635J+7.2J\nE_(t)=8.4635J$

If the radio waves transmitted by a radio station have a frequency of 83.5 MHz, what is the wavelength of the waves, in meters

Answers

Answer: wavelength =3.52m

Explanation:

,λ=c/μ

where c=speed of the light,λ=wave length, μ=frequncy

c=3x10^8m/s

And

μ=83.5/MHz =85.3x10^6Hz==85.3x10^6Hz=

=85.3x10^6s-1

λ=c/μ

=3x10^8m/s/85.3x10^6s-1

=3.51699883

=3.52m

Assume we are given an electric field set up by an unknown charge distribution. U0 is the amount of work needed to bring a point charge of charge q0 in from infinity to a point P. If the charge q0 is returned to infinity, how much work would it take to bring a new charge of 4 q0 from infinity to point P?

Answers

Answer:

$4U_0$

Explanation:

We are given that

Amount of work needed to bring a point charge q0 from infinity to a point P= $U_0$

We know that potential at point P= $V=(U)/(q)$

$U=Vq$

Where U=Amount of work needed to bring a point charge q from infinity to a point P

Initially , $V=(U_0)/(q_0)$

New charge, q= $4q_0$

Then, work done,U= $(U_0)/(q_0)* (4q_0)=4U_0$

Hence, the amount of work needed to bring a new charge 4q0 from infinity to point P= $4U_0$

Answer:

$4U_(0)$

Explanation:

V = u / q,

Work = P = V

1U / 1/4 = 4U

Answer plz answer plzzz I am a little confused with full time

Answers

I can’t read that I’m sorry make it more clear

A circular loop of wire has radius 7.80cm . A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 2.03�10?2W/m2 , and the wavelength of the wave is 6.20m .What is the maximum emf induced in the loop? Express your answer with the appropriate units.I stumbled through the formulas I do know for EMF but cant seem to figure out how to get the right answer. Please help and provide explanation! Thanks

Answers

Answer:

fem = - 4.50 10²² V

Explanation:

For the solution of this problem we must use the equation of the induced electromotive force or Faraday's law

E = - d Φ $._(B)$ / dt = d (BA cos θ) dt

In this case they tell us that the magnetic field is perpendicular to the plane of the loop, as the normal to the surface of the loop is in the direction of the radius, the angle between the field and this normal is zero, so cos 0º = 1. The area of the loop is constant, with this the equation is

E = - A dB / dt (1)

To find field B, we have the relationships of electromagnetic waves

E = c B

The intensity or poynting vector for the wave is described by the equation

S = I = 1 / μ₀ E x B = 1 /μ₀ E B

We replace

I = 1 /μ₀ (cB) B = c /μ₀ B²

This is the instantaneous intensity.

B = √ (μ₀ I /c)

We substitute in equation 1

E = - A μ₀/c d I / dt

With the maximum value we are asked to change it derived from variations

E = -A c/μ₀ ΔI / Δt

It remains to find the time of the variation. Let's use the equation

c = λ f = λ / T

T = λ / c

T = 6.20 / 3 10⁸

T = 2.06 10⁻⁸ s

We already have all the values to calculate the fem

fem = - π r² c/μ₀ ΔI/Δt

fem = - (π 0.078²) (3 10⁸/(4π 10⁻⁷) (2.03 10² -0) / (2.06 10⁻⁸ - 0)

fem = - 4.50 10²² V

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An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

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