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Westkost [7]
3 years ago
9

A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor

has inductance 1.82 mH, then what is the required capacitance of the capacitor? Select one: a. 8.3e-4 Farads b.0.12 Farads c. 6.8e-10 Farads d. 2.7e-5 Farads e. 6.8e-7 Farads
Physics
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

(e)6.835\times 10^{-7}F

Explanation:

At resonance we know that X_l=X_C

That is \omega L=\frac{1}{\omega C}

\omega ^2=\frac{1}{LC}

\omega =\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So 4511=\frac{1}{2\pi \sqrt{LC}}

LC=\frac{1}{4\pi ^2\times 4511^2}

LC=1.244\times 10^{-9}

C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F

So option e is the correct answer

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The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
A standard steel pipe (D = 3.81 in.; d = 3.24 in.) supports a concentrated load of P = 930 lb. The span length of the cantilever
Maurinko [17]

Answer:

a) τmax = 586.78 P.S.I.

b) σmax = 15942.23 P.S.I

Explanation:

D = 3.81 in

d = 3.24 in

P = 930 lb

L = 3.7 ft = 44.4 in

a) The maximum horizontal shear stress can be obtained as follows

τ = V*Q / (t*I)

where

V = P = 930 lb

Q = (2/3)*(R³- r³) = (1/12)*(D³- d³) = (1/12)*((3.81 in)³- (3.24 in)³)

⇒ Q = 1.7745 in³

t = D - d = 3.81 in - 3.24 in = 0.57 in

I = (π/64)*(D⁴-d⁴) = (π/64)*((3.81 in)⁴- (3.24 in)⁴) = 4.9341 in⁴

then

τ = (930 lb)*(1.7745 in³) / (0.57 in*4.9341 in⁴)

⇒ τmax = 586.78 P.S.I.

b) We can apply the following equation in order to get the maximum tension bending stress in the pipe

σmax = Mmax *y / I

where

Mmax = P*L = 930 lb*44.4 in = 41292 lb-in

y = D/2 = 3.81 in /2 = 1.905 in

I = 4.9341 in⁴

then

σmax = (41292 lb-in)*(1.905 in) / (4.9341 in⁴) = 15942.23 P.S.I

6 0
4 years ago
How much would a 25 kg suitcase weigh on the surface of…?
photoshop1234 [79]

Answer:

A. 95N

B. ?

C.225N

D.?

Explanation:

For Mars and Pluto im not so sure about those too but for A and C I am positive those are correct.

Sorry I could not help you all the way, please dont be mad ;(

8 0
3 years ago
A father pushes his child in a cart. The cart starts to move.
LuckyWell [14K]

The potential energy of the car when it let go is 20,000 J.

The speed of the car at the bottom of the ramp is 20 m/s.

The given parameters;

  • <em>mass of the car, m = 100 kg</em>
  • <em>height of the car, h = 20 m</em>

<em />

The potential energy of the car is calculated as follows;

P.E = mgh

P.E = 100 x 10 x 20

P.E = 20,000 J

The speed of the car at the bottom of the ramp is calculated as follows;

K.E = P.E\\\\\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2 \times 10 \times 20} \\\\v = 20 \ m/s

Learn more here:brainly.com/question/18597080

5 0
3 years ago
A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca
natka813 [3]

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

7 0
4 years ago
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