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3 years ago

If the frequency of wave is 400 Hz and its

Physics

1 answer:

agasfer [191]3 years ago

8 0

Answer:

D) 1000 m/s

Explanation:

$wavelength = \frac{velocity}{frequency}$

You have the wavelength and frequency, you just need to solve for velocity. You can do this by multiplying each side of the equation by frequency.

Hope this helps.

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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

Read 2 more answers

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur

Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v $_r$ = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

v $_r$ = ( u + u ) / ( 1 + ( u×u / c²) )

v $_r$ = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )

we substitute

v $_r$ = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )

v $_r$ = 1.904c / ( 1 + 0.906304 )

v $_r$ = 1.904c / 1.906304

v $_r$ = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0

3 years ago

HELP Giving all the points I can. Physics virtual lab report: circuit design<br><br> Please help me.

mafiozo [28]

Answer:text me I can help

Explanation:

5 0

3 years ago

Read 2 more answers

Two asteroids are 75000 m apart one has a mass of 8 x 10^7 kg if the force of gravity between them is 1.14 what is the mass of t

vaieri [72.5K]

I’m going to say B

I hope this helped

4 0

3 years ago

Read 2 more answers

Gaining neutrons makes an element?

Kobotan [32]

Answer:

C. Have no change in electrical charge

Explanation:

If a element gains neutron it become an Isotope. The electrical charge do not change for this, only the atomic mass changes when an element gains neutrons.

The electrical charge is affected when there is a variation in the number of electrons or protons in the element.

6 0

3 years ago