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kati45 [8]
3 years ago
13

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde

nly applied to the initially uncharged plates through a 1000 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 50 volts?
Physics
1 answer:
Afina-wow [57]3 years ago
8 0

Answer:Resistor-Capacitor (RC) circuits, when driven by a voltage/current source, display a type of time-dependent charging or discharging since the charge from the capacitor goes through the resistor. When considering a discharging phase, the time-dependent voltage takes the form

Explanation:

Vt = Voe^-t/RC

Vt equal time dependent voltage

R is the resistance

C is the capacitance of the oscilloscope

Given

Area of capacitor equal to 0.1 * 0.002m

Distance between plates equal 1/1000 = 0.0001m

Vo = 100v supply voltage

Resistance R =:1000ohms

Vt time dependent voltage = 50V

To find capacitance C = Eo(A/d)

C = (8.85 * 10^-12)* (0.002/0.001)

C= 1.77 * 10^-11

Solving the equation Vt = Voe^-t/RC

t = 0.6931RC

t = 0.6931(1000)(1.77*10^-11)

t = 1.2*10^-9

t = 1.2ns

It will take the oscilloscope 1.2ns to reach 50V

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A motor exerts a force of 12,00N to lift an elevator 8.0m in 7.0secs what is the power produced by the motor?
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A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
3 years ago
URGENT!! <br>Question in attachment. <br>thanks ​
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Answer:9

Explanation:beacause

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Kyra and Pari are timing how long it takes for 1 g of sugar to dissolve in hot water. Kyra records a time of 24.3 seconds. Pari
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<h3>Answer:</h3><h3>we can say that:-</h3>
  1. A reading with more no of significant figures is considered to be more precise.
  2. Kyra recorded a reading of 24.3 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
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