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1 year ago

6

In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ

ency is of particular interest. On the basis of the ICNIRP guidelines, what is the maximum intensity of an electromagnetic wave at this frequency to which the general public should be exposed

1 answer:

grigory [225]1 year ago

7 0

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E $_{max$ is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E $_{max$ )

E $_{max$ = 350/f kV/m

given that frequency of 60 Hz, we substitute

E $_{max$ = 350/60 kV/m

E $_{max$ = 5.83333 kV/m

E $_{max$ = 5.83333 kV/m × ( $\frac{1000 V/m}{1 kV/m}$ )

E $_{max$ = 5833.33 N/C

so we substitute all our values into the formula for intensity of an electromagnetic wave;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

$I$ = $\frac{1}{2}$ × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

$I$ = 45 × 10³ W/m²

$I$ = 45 × 10³ W/m² × ( $\frac{1 kW/m^2}{10^3W/m^2}$ )

$I$ = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

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Andreas93 [3]

Answer:

a) W = 643.5 J, b) W = -427.4 J

Explanation:

a) Work is defined by

W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

W = F x

let's calculate

W = 165 3.9

W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

sin θ = Wₓ / W

cos θ = Wy / W

Wₓ = W sinθ = mg sin θ

Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

W = Wₓ x cos 180

W = - mg sin 34 x

let's calculate

W = -20 9.8 sin 34 3.9

W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

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Determine the correlation between coronal mass ejections from the Sun to the accumulation of the rare and valuable isotope He3 t

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2 years ago

You hang an object with mass m = 0.380 kg from a vertical spring that has negligible mass and force constant k = 60.0 N/m. You p

joja [24]

Answer:

a) 0.500 s

b) greater than 0.500 s

c) greater than 0.500 s

Explanation:

The time period of an oscillating spring-mass system is given by:

$T=2\pi \sqrt{\frac{m}{k}}$

where, m is the mass and k is the spring constant.

a) As the period of oscillation does not depend on the distance by which the mass is pulled, the period would remain same as 0.500 s for the given system.

b) As the period varies inversely with the square root of spring constant, so with the decrease in the spring constant, the period would increase. So, the new period would be greater than 0.500 s.

c) As the period varies directly with the square root of mass, so with the increase in mass, the period will also increase. The new period will be greater than 0.500 s.

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2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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2 years ago

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lianna [129]

Answer:

$\lim_{t \to \infty} v(t) =vT$

Explanation:

Using distributive propierty:

$v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}$

So:

$\lim_{t\to \infty} vT-e^{-gt}$

The limit of the sum of two functions is equal to the sum of their limits, therefore:

$\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}$

The limit of a constant function is the constant, hence:

$\lim_{t\to \infty} vT=vT$

Now, let's solve the other limit:

$\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}$

The limit of a constant times a function is equal to the product of the constant and the limit of the function, so:

$\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)$

$-g(\infty)=-\infty$

Therefore:

$e^{(-\infty)} =0$

Finally:

$\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT$

8 0

2 years ago