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daser333 [38]
3 years ago
6

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

w many time constants (a decimal number) must elapse before a capacitor in a series RC circuit is charged to 39.0% of its equilibrium charge
Physics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer: t = 0.492τ

Explanation: In a circuit where there is a resistor and a capacitor, the equation for a charging capacitor is given by:

q = q_0(1 - e^{\frac{t}{RC} })

where:

q_0 is the equilibrium charge

q is the charge at time t

RC is time constant also called τ (tau)

For this problem, the circuit is charged to 39%, which means: q = 0.39 q_0

0.39q_0 = q_0 (1 - e^\frac{-t}{RC} )

0.39 = 1 - e^\frac{-t}{RC}

e^\frac{-t}{RC} = 0.61

\frac{-t}{RC} = ln(0.61)

-t = ln(0.61)τ

t = 0.492τ

For the condition to be met it is needed 0.492 time constants must elapse.

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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
4 years ago
What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed
Pani-rosa [81]

     charge must be equal to 5.74 ×10⁻⁵

 In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

     →    Fnet =0

     →    mg =  qE

 substituting the values we get :

         0.00345 × 9.81 =  q × 590

   →       q = 5.74 ×10⁻⁵

    Hence the charge must be equal to   5.74 ×10⁻⁵.

   Learn more about charges here:

          brainly.com/question/26092261

                    # SPJ4

8 0
2 years ago
An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal inci
VikaD [51]

Answer:

572.3 nm

Explanation:

n_{oil} = refractive index of the oil film = 1.48

t_{oil} = thickness of the oil film = 290 nm

\lambda = wavelength of the dominant color

m = order

Using the equation

2 n_{oil} t_{oil} = (m + 0.5) \lambda

For m = 0

2 (1.48) (290) = (0 + 0.5) \lambda

\lambda = 1716.8 nm

For m = 1

2 (1.48) (290) = (1 + 0.5) \lambda

\lambda = 572.3 nm

For m = 2

2 (1.48) (290) = (2 + 0.5) \lambda

\lambda = 343.4 nm

Hence the dominant color wavelength is 572.3 nm

8 0
4 years ago
Please HELP!!
Rudik [331]

Answer:

I think is 2.

Explanation:

(The entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light)

7 0
3 years ago
Hot coffee in a mug cools over time and the mug warms up. Which describes the energy in this system?
givi [52]
Thermal energy from the coffee is transferred to the mug.
3 0
3 years ago
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