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2 years ago

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Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a

larm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and capacitor in series. The maximum voltage across the capacitor is to be 12.0 V . To produce a sufficiently loud sound, the capacitor must store an amount of energy equal to 1.55×10−2 J . What values of capacitance and inductance should you choose for your car-alarm circuit?

1 answer:

Alex17521 [72]2 years ago

6 0

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

$V =12.0 V$ , $E = 1.55*10^2 J$ , $f = 3600 Hz$

To determine the capacitance can use the equation

$U_c= \frac{1}{2}*C*V^2$

Solve to C'

$C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}$

$C=215.27 uF$

To find the inductance can use the frequency of the circuit

$f = \frac{1}{2\pi* \sqrt{C*L} }$

Solve to L'

$L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}$

$L = 9.078 uH$

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A 18.0−μF capacitor is placed across a 22.5−V battery for a few seconds and is then connected across a 12.0−mH inductor that has

Gelneren [198K]

Answer:

Part a)

$i = 10.4 mA$

Part b)

in this case the charge on the capacitor will become zero

Part c)

$t_1 = 0.73 ms$

Part d)

$t = 2.2 ms$

Explanation:

As we know that first capacitor is charged with the battery and then it is connected to the inductor

So here we will have

$Q = CV$

$Q = (18\mu F)(22.5 V)$

$Q = 405 \mu C$

Part a)

now since the total energy of capacitor is converted into the energy of inductor

so by energy conservation we can say

$\frac{Q^2}{2C} = \frac{1}{2}Li^2$

so maximum current is given as

$i = \sqrt{\frac{L}{C}}Q$

$i = \sqrt{\frac{12\times 10^{-3}}{18\times 10^{-6}}}(405\times 10^{-6})$

$i = 10.4 mA$

Part b)

When current is maximum then whole energy of capacitor is converted into magnetic energy of inductor

So in this case the charge on the capacitor will become zero

Part c)

Time period of oscillation of charge between the plates and inductor is given as

$T = 2\pi\sqrt{LC}$

$T = 2\pi\sqrt{(18\mu F)(12 mH)}$

$T = 2.92 ms$

now capacitor gets discharged first time after 1/4 of total time period

$t_1 = 0.73 ms$

Part d)

Since time period is T and capacitor gets discharged two times in one complete time period of the motion

so first it will discharges in T/4 time

then next T/4 it will get charged again

then next T/4 time it will again discharged

so total time taken

$t = 3T/4$

$t = 2.2 ms$

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2 years ago

For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?

Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

$85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}$

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity ( $v_t$ ), in order to obtain the angular velocity of the wheel:

$v_t=w*R\\w=\frac{v_t}{R}$

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

$w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}$

And now, recalling that $2\pi$ radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by $2\pi$ :

rotational frequency = $\frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}$

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2 years ago

An electron is trapped in a square well of unknown width, L. It starts in unknown energy level, n. When it falls to level n-1 it

Lady bird [3.3K]

Answer:

(1) En to n-1 = 0.55 ev

(2) En-1 to n-2 = 0.389 ev

(3) ninitial =4

(4) L =483.676 ×10^-11 nm

(5) λlongest= 1773.33 nm

Explanation:

Detailed explanation of answer is given in the attached files.

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2 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

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2 years ago

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscil

marishachu [46]

Answer:

a) 40,000 N/m

b) f = 6.37 Hz

c) v = 4,8 m/s

Explanation:

part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

k = 3200 / 0.02 = 160,000 N/m

This means that one spring will be:

k' = 160,000 / 4 = 40,000 N/m

b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

k^n = 2k

To get the frequency, we need to know the angular speed of the car with the following expression:

wo = √k^n / M

M: mass of the wheel and axle, which is 50 kg

k = 40,000 N/m

Replacing the data:

wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

x: distance of 80 cm or 0.8 m

V = 0.8 * 6 =

V = 4.8 m/s

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2 years ago