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3 years ago

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You have an RC timing circuit made from three capacitors, each C and in parallel with each other, and one resistor R. You wantto

change the RC time constant to one-third what it was byremoving one of the capacitors and adding one resistor of theappropriate value. What does the new total resistance of theciruit need to be and how should you add the new resistor (seriesor parallel)

1 answer:

nika2105 [10]3 years ago

8 0

Answer:

New total resistance = R/2

The new resistor is connected in parallel.

Explanation:

Let $T_1$ be initial time constant and $T_2$ the new time constant.

The 3 capacitors are connected in parallel. The total capacitance, $C_t$ , is their sum. Hence

$C_t=C + C + C = 3C$

$T_1 = R \times 3C = 3RC$

The new time constant is one-third of the initial time constant.

$T_2 = \dfrac{T_1}{3} = RC$

If one of the capacitors is removed, then the new total capacitance, $C_{t2}$ is

$C_{t2} = 2C$

If the new total resistance isc $R_{t2}$ , then

$T_2 = R_{t2}C_{t2}$

$RC = R_{t2}\times2C$

$R_{t2} = \dfrac{R}{2}$

Since this is less than the old total resistance, the new resistor, with resistance, X, must be connected in parallel. Its value will be R, as below:

$\dfrac{1}{R/2}=\dfrac{1}{R}\dfrac{1}{X}$

$\dfrac{2}{R}-\dfrac{1}{R}=\dfrac{1}{X}$

$\dfrac{1}{R}=\dfrac{1}{X}$

$X = R$

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A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take

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Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$x_{o}$ , $y_{o}$ - Initial horizontal and vertical position of the projectile, measured in meters.

$v_{o,x}$ , $v_{o,y}$ - Initial horizontal and vertical speed of the projectile, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

$x$ , $y$ - Current horizontal and vertical position of the projectile, measured in meters.

Given that $x_{o} = 0\,m$ , $y_{o} = 80\,m$ , $v_{o,x} = 360\,\frac{m}{s}$ , $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the kinematic equations are, respectively:

$x = 360\cdot t$

$y = 80-4.094\cdot t^{2}$

(a) If $y = 0\,m$ , the time taken for the projectile to reach the water is:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If $t \approx 4.420\,s$ , the horizontal scope of the projectile is:

$x = 360\cdot (4.420)$

$x = 1591.2\,m$

The horizontal scope of the projectile is 1591.2 meters.

(c) If $t = 2\,s$ , the height that remains to descend is:

$y = 80-4.094\cdot (2)^{2}$

$y = 63.624\,m$

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

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Explanation:

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= 3.0 $ft^{3}$

As mass remains constant then the specific volume at this state will be as follows.

$\nu_{2} = \frac{V_{2}}{m}$

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$x_{2}$ = 0.229

Now, the final pressure will be the saturation pressure at $T_{2}$ = 300 F

and, $P_{2}$ = $P_{sat}$ = 66.985 psia

Formula to calculate internal energy at the final state is as follows.

$U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}$

= $2(269.51 + 0.229 \times 830.45)$

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