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2 years ago

8

A gas-powered lawnmower is rated at 98 db. how many times more intense is the sound of this mower than that of an electric-power

ed mower rated at 70 db?

1 answer:

givi [52]2 years ago

6 0

The gas- powered mower is

(98dB - 70dB) = 28 dB louder.

28 dB = 10^2.8

= 631 times as much sound power.

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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

Stress distributed over an area is best described as: a) External force b) Axial force c) Radial force d) Internal resistive for

Anit [1.1K]

Answer:

Option D is the correct answer.

Explanation:

Stress is the force per unit area that tend to change the shape of body.

Stress is defined as internal resistive force per unit area.

$\texttt{Stress}=\frac{\texttt{Internal resistive force}}{\texttt{Area}}$

$\sigma =\frac{F}{A}$

So, so stress distributed over an area is best described as internal resistive force.

Option D is the correct answer.

8 0

3 years ago

Sino tao ang gutom?<br>A.you<br>B.mama mo!<br>C.tamod

Fiesta28 [93]

The answer to that question is c. tamod

5 0

2 years ago

A lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms , what is the average

malfutka [58]

The average electric current in the lightning will be 8 × $10^{-17}$ A

<h3>Why Lightning Conductors on top of a tower ?</h3>

The lightning conductors are long metal strips running from the spike end of a conductor on the top of a building to the earth. They are used to prevent buildings from destruction when struck by thunder or lightning.

Given that a lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms, to calculate the average electric current in the lightning, we will first consider the charge released.

one charge = 1.6 × $10^{-19}$ C

Average current I = Q/t

Where

Q = charge = 1.6 × $10^{-19}$ C

t = time = 2ms = 2 × $10^{-3}$ s

I = current = ?

Substitute all the parameters into the formula

I = 1.6 × $10^{-19}$ C ÷ 2 × $10^{-3}$

I = 8 × $10^{-17}$ A

Therefore, the average electric current in the lightning will be 8 × $10^{-17}$ A

Learn more about Lightning here: brainly.com/question/3183045

#SPJ1

5 0

2 years ago

How many neutrons does element X have if its atomic number is 22 and its mass number is 96?

Dovator [93]

1. The mass number is protons + neutrons = mass number. In this case, we have protons + neutron = 164.The atomic number is simply the number of protons so we have 43 + neutrons = 164. Subtracting 43 from both sides we get neutrons = 121.
2. = 4

3. The number of protons in the nucleus does not equal the number of neutrons.

A=mass number:
Z=atomic number (= number of protons)
N=number of neutrons:

A=Z+N

If the number of protons in the nucleus is equal the number of neutrons , we would have an even mass number; because Z=N=x; then A=x+x=2x (this is always an even number) but 23 is an odd number, therefore the number of protons in the nucleus does not equal the number of neutrons.

6 0

2 years ago