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3 years ago

5

5. __________ is the idea that all people must follow the law and that the law is applied to all equally even people in the gove

rnment.
Public Policy
Rule of Law
General Welfare
Natural Rights

1 answer:

padilas [110]3 years ago

5 0

The answer is Rule of the law

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WGVU-AM is a radio station that serves the Grand Rapids, Michigan area. The main broadcast frequency is 1480kHz. At a certain di

Yuki888 [10]

Answer:

a

$\lambda = 202.7 \ m$

b

$w = 9.3 *10^{6} \ rad/s$

c

$k = 0.031 m^{-1}$

d

$E_{max} = 9.0 *10^{-3} \ V/m$

Explanation:

From the question we are told that

The frequency of the radio station is $f= 1480 \ kHz = 1480 *10^{3}\ Hz$

The magnitude of the magnetic field is $B = 3.0* 10^{-11} \ T$

Generally the wavelength is mathematically represented as

$\lambda = \frac{c}{f}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$\lambda = \frac{3.0 *10^{8}}{ 1480 *10^{3}}$

=> $\lambda = 202.7 \ m$

Generally the angular frequency is mathematically represented as

$w = 2 \pi * f$

=> $w = 2 * 3.142 * 1480 *10^{3}$

=> $w = 9.3 *10^{6} \ rad/s$

Generally the wave number is mathematically represented as

=> $k = \frac{2 \pi }{\lambda}$

=> $k = \frac{2 * 3.142 }{ 202.7}$

=> $k = 0.031 m^{-1}$

Generally the amplitude of the electric field at this distance from the transmitter is mathematically represented as

$E_{max} = c * B$

=> $E_{max} = 3.0 *10^{8} * 3.0* 10^{-11}$

=> $E_{max} = 9.0 *10^{-3} \ V/m$

3 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

LekaFEV [45]

Answer:

(h₁-h₂) = 2.30 10² m

Explanation:

The pressure depends on the height with the formula

P = P_atm + rho g h

Let's apply this expression for the building

P₁ = P_atm + rho_air g h₁

P₂ = P_atm + rho_air g h₂

Subtract

P₁ - P₂ = roh_air g (h₁ –h₂)

The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity

P₁- P₂ = rho_Hg g (h₁-h₂) _Hg

rho_Hg g (h₁-h₂) _Hg = roh_air g (h₁ –h₂)

(h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg

Let's calculate

(h₁-h₂) = 13600 / 1.18 (695-675)

(h₁-h₂) = 2.30 10⁵ mm

Let's reduce to meter

(h₁-h₂) = 2.30 10⁵ mm (1 m / 10³ mm)

(h₁-h₂) = 2.30 10² m

4 0

3 years ago

All celestial objects we tracted to each other by gravity.Asteroids A and B are 5 meters (m) part. Asteroids C and D are also 5

Nadusha1986 [10]

A. Between asteroids C and D because gravitational attraction is greatest with increased mass.

6 0

3 years ago

Water runs out of a horizontal drainpipe at the rate of 135 kg/min. It falls 3.1 m to the ground. Assuming the water doesn't spl

Arlecino [84]

Answer:

The average force exerted by the water on the ground is 17.53 N.

Explanation:

Given;

mass flow rate of the water, m' = 135 kg/min

height of fall of the water, h = 3.1 m

the time taken for the water to fall to the ground;

$h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s$

mass of the water;

$m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg$

the average force exerted by the water on the ground;

F = mg

F = 1.789 x 9.8

F = 17.53 N

Therefore, the average force exerted by the water on the ground is 17.53 N.

7 0

3 years ago

Explain briefly why the intensity reflected off the back surface of the film (i.e., the right surface, where there is a liquid-t

Katarina [22]

Explanation:

Taking the incident light to be traveling in the + x-direction so that it is at normal incidence to the left side of the film(referred to as the "Front side"). This means the beam transmitted into the liquid is essentially as strong as the incident beam.

Almost all the light that is reflected off the back surface will get through the front surface. (But only 2.78% gets re-reflected off the the front surface back to the right) this means that there are two beams reflected to the - x-direction, one from the front surface and one from the back, and these beams are of almost equal intensity.

Hope this is helpful. Thanks

7 0

3 years ago