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Gnoma [55]
3 years ago
9

Now Ishani and John are ready to use what they know to build the turn indicator. They decide that the signal in their cars goes

about once per second. Assuming that they set the trigger to make the indicator go when charge has reached 1 − 1/e of its final value, and that it takes about the same amount of time to discharge as to charge, suggest values for R and C that could be used. (Assume that the capacitor reaches approximately its maximum value, discharges to 1 − 1/e of its maximum value over the course of 1 second, and then begins charging again.)'
Physics
1 answer:
Blababa [14]3 years ago
8 0

Answer:

For a resistance (R) of 1000 MΩ and capacitance (C) of 2.179x10⁻³μF

Explanation:

The charge on the discharging capacitor is equal:

Q=Q_{o} e^{-\frac{t}{RC} }

If t = 1 s, then:

Q=Q_{o} (1-\frac{1}{e} )

Matching both equations:

Q_{o} e^{-\frac{t}{RC} } =Q_{o}( 1-\frac{1}{e})\\e^{-\frac{t}{RC} } =1-\frac{1}{e}

If t = 1

e^{-\frac{1}{RC} } = 1-\frac{1}{e}\\e^{-\frac{1}{RC} } =0.632\\-\frac{1}{RC} =ln0.632\\-\frac{1}{RC}=-0.459\\RC=\frac{1}{0.459} =2.179ohmF

The capacitance for a resistor of resistance of 1000 MΩ is equal:

C=\frac{2.179}{R} =\frac{2.179}{1000x10^{6} } =2.179x10^{-9} =2.179x10^{-3} \mu F

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A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3
Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0
3 years ago
The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c
Natalka [10]

Answer:

EMF = 1684.67 Volts

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

EMF = \frac{d\phi}{dt}

now we have

\phi = BA

here since magnetic field is constant so we have

EMF = A\frac{dB}{dt}

now we have

A = (190 \times 10^3)(190 \times 10^3)

A = 3.61 \times 10^{10} m^2

now we have

EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})

EMF = 1684.67 Volts

6 0
3 years ago
a wave having a wavelength of 4.7 meters and an amplitude of 2.5 meters travels a distance of 28 meters in 7 seconds. determine
Aleksandr [31]

Answer:

<h2>T(Period) = 1.33s</h2><h2>f(Frequency) = 0.75Hz (cycles/second)</h2>

Explanation:

<h2>Given:</h2><h2 /><h2>λ = 4.0m</h2><h2>Amplitude = 25m</h2><h2>d = 24m</h2><h2>s = 8.0s</h2><h2 /><h2>Required:</h2><h2>f = ?</h2><h2>T = ?</h2><h2 /><h2>Analysis:</h2><h2>v = λf</h2><h2>f =N/t</h2><h2>T = 1/f</h2><h2 /><h2>v = d/t</h2><h2 /><h2>Solve:</h2><h2>v = d/t = 24/8.0 → v = 3.0m/s</h2><h2>v =λf → f = v/λ = 3.0/4.0 → f = 0.75Hz</h2><h2>T = 1/f = 1/0.75 → T = 1.33s</h2><h2 /><h2>Hopes this helps. Mark as brainlest plz!</h2>
4 0
3 years ago
A dart is thrown horizontally at a target's center that is 5.00 m, away. The dart hits the target 0.150 m below the targets cent
boyakko [2]

Answer:

  v₀ₓ = 28.6 m / s

Explanation:

This is a missile throwing exercise

          y = v_{oy} t - ½ g t²

as the dart is thrown horizontally the vertical velocity is zero (I go = 0)

          y = - ½ g t²

          t = \sqrt{- \frac{2y}{g}  }

let's calculate

         t = \sqrt{- \frac{2 \  (-0.150)}{9.8} }

         t = 0.175 s

the expression for horizontal displacement is

         x = v₀ₓ t

         v₀ₓ = x / t

         v₀ₓ = 5.00 / 0.175

         v₀ₓ = 28.6 m / s

7 0
3 years ago
The Sun and other stars orbit Earth.
ratelena [41]
What is the question?
7 0
3 years ago
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