Um aluno pretende construir um aquecedor usando um enrolamento de fio. Experimenta e verifica que não

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6

lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

The amplitude of induced emf by this signal between the ends of the receiving antenna is

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

Here,

I = Current

$\mu_0$ = Permeability at free space

c = Light speed

d = Distance

Replacing,

$\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}$

$\epsilon_{max} = 0.05434V$

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0

3 years ago

A box sits at rest on a rough 33° inclined plane. Draw the free-body diagram, showing all the forces

Cloud [144]

(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

<h3>Free body diagram</h3>

The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;

/ W2

Ф → Ff

↓W1

where;

Ff is the frictional force resisting the down motion of the box
W1 is the perpendicular component of the box weight = Wcos(33)
W2 is the parallel component of the box weight = Wsin(33)

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

Learn more about free body diagram of inclined objects here: brainly.com/question/4176810

8 0

2 years ago

Rank the five terrestrial worlds in order of size from smallest to largest: Group of answer choices Mercury, Venus, Earth, Moon,

dimaraw [331]

Answer: Moon, Mercury, Mars, Venus, Earth

Explanation:

7 0

2 years ago

Mr. Phillips' car is parked on a steep hill with the brakes applied and the engine off. Because of the car's position, it has gr

aivan3 [116]

Answer:

The anser is b

Explanation:

4 0

3 years ago

The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space

Svet_ta [14]

D not 100% sure though.

4 0

3 years ago