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2 years ago

12

60 POINTS model how sound produced by a singer is recorded by a microphone and reproduced by a speaker (sound model) WITH RUBIC

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1 answer:

Nikolay [14]2 years ago

6 0

Answer:

Explanation: i don't know bro

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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

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

julia-pushkina [17]

Answer:

$5.5\cdot 10^{-11} C$

Explanation:

The capacitance of the parallel-plate capacitor is given by:

$C=\epsilon_0 \frac{A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=190 mm^2 = 190 \cdot 10^{-6} m^2$ is the area of the plates

$d=1.2 mm = 0.0012 m$ is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F$

The energy stored in the capacitor is given by

$U=\frac{Q^2}{2C}$

Since we know the energy

$U=1.1 nJ = 1.1 \cdot 10^{-9} J$

we can re-arrange the formula to find the charge, Q:

$Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C$

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2 years ago

A vector must always have both size and ..

DochEvi [55]

Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up

5 0

3 years ago

How do you solving kinematic equations for horizontal projectiles?

daser333 [38]

See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION

7 0

2 years ago

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.

sineoko [7]

Answer:10842.33m/s

Explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s

5 0

3 years ago