(0 10 40 20 30 50 70 80) (90 100 110 120 130 140). Which end of the scale (left or right) has the loudest sounds?

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

4 years ago

Read 2 more answers

The transfer of energy by the movement of particles that are in contact with each other

jeka57 [31]

Answer: I belive that the Answer is C.) Conduction

Explanation:

3 0

3 years ago

Read 2 more answers

No links please, links never work, this is science not physics

stiks02 [169]

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

7 0

3 years ago

Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m^2. Assuming that all the photons in th

andrew11 [14]

Answer:

3.066×10^21 photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

= 3.066×10^21

Therefore, the number of photons per second per square meter 3.066×10^21 photons/(s.m^2).

4 0

3 years ago

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a

aleksklad [387]

Answer:

a) y = 0.98 t², t=1s y= 0.98 m,

b) he two blocks must move the same distance

c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

N-W = 0

N = M g

X axis

T- fr = Ma

the friction force has the expression

fr = μ N

fr = μ Mg

small block

w- T = m a

we write the system of equations

T - fr = M a

mg - T = m a

we add and resolved

mg- μ Mg = (M + m) a

a = $g \ \frac{m - \mu M}{m+M}$

a = $9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}$

a = 9.8 (6/30)

a = 1.96 m / s²

a) now we can use the kinematic relations

y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

y = ½ a t²

y = ½ 1.96 t²

y = 0.98 t²

for t = 1s y = 0.98 m

t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

v = vo + a t

v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

-fr = M a

- μ M g = M a

a = - μ g

a = - 0.2 9.8

a = -1.96 m / s²

e) the distance to stop the block is

v² = vo² - 2 a x

0 = vo² - 2a x

x = vo² / 2a

x = 1.96² / 2 1.96

x = 0.98 m

the time to travel this distance is

v = vo - a t

t = vo / a

t = 1.96 /1.96

t = 1 s

3 0

3 years ago

(0 10 40 20 30 50 70 80) (90 100 110 120 130 140). Which end of the scale (left or right) has the loudest sounds?​

(0 10 40 20 30 50 70 80) (90 100 110 120 130 140). Which end of the scale (left or right) has the loudest sounds?