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bazaltina [42]
2 years ago
10

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?
Physics
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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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0
2 years ago
The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

B=\frac{\mu _o I}{2\pi r}

where B= magnetic field

           \mu _o= permeability of free space

           I= current in the long wire and

           r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

   

7 0
2 years ago
Read 2 more answers
Consider a string of length 1.0 meter, fixed at both ends, with mass 100 grams and tension 100 newtons. part a give the number o
Bond [772]
To answer the problem we would be using this formula which isv = sqrt(T/(m/L)) 
v = sqrt(100 N / [(0.100 kg)/(1.0 m)]) 
v = 31.62 m/s 
v = fλ 
31.62 m/s = (95 Hz)(λ) 
λ = 0.333 m 
For every wavelength along a string there will be 2 antinodes. 
1.0 m / 0.333 m = 3 
3 * 2 = 6 antinodes 
6 + 1 = 7 nodes
4 0
2 years ago
Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a
serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

\omega = 1.25 rad/s \rightarrow The angular speed

\alpha = 0.745 rad/s2 \rightarrow The angular acceleration

r = 4.65 m \rightarrow The distance

The relation between the linear velocity and angular velocity is

v = r\omega

Where,

r = Radius

\omega = Angular velocity

At the same time we have that the centripetal acceleration is

a_c = \frac{v^2}{r}

a_c = \frac{(r\omega)^2}{r}

a_c = \frac{r^2\omega^2}{r}

a_c = r \omega^2

a_c = (4.65 )(1.25 rad/s)^2

a_c = 7.265625 m/s^2

Now the tangential acceleration is given as,

a_t = \alpha r

Here,

\alpha = Angular acceleration

r = Radius

\alpha = (0.745)(4.65)

\alpha = 3.46425 m/s^2

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

|a| = \sqrt{a_c^2+a_t^2}

|a| = \sqrt{(7.265625)^2+(3.46425)^2}

|a| = 8.049 m/s^2 \approx 8.05 m/s2

Therefore the correct answer is C.

7 0
2 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
Oksana_A [137]
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>

Explanation:

When the charge of 2e is placed in between the plates .

The force applied on this charge by plates is = q E

here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C

and E is the magnitude of electric field intensity

The work done = Force x displacement

Thus W = q E x S

here S is displacement

Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8

= 1.02 x 10⁻¹⁷ J

This work will be converted into the kinetic energy of charge .

Thus K.E = 1.02 x 10⁻¹⁷ J

3 0
2 years ago
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