Ask question

Ask question

All categories

3 years ago

15

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

usoidal voltage with an amplitude of 13.0 V is applied across the inductor.What frequency is required?

1 answer:

faust18 [17]3 years ago

6 0

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$

Current in the inductor $i=1.90mA=1.90\times 10^{-3}A$

Voltage v = 13 volt

Inductive reactance of the circuit $X_l=\frac{v}{i}$

$X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm$

We know that

$X_l=\omega L=2\pi fL$

$2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10$

f = 2421.127 kHz

You might be interested in

Read the information and study the image.

erastova [34]

Answer:

It is impossible to detect underground water from the surface. Dowsing practitioners refuse to explain their secrets.

Explanation:

4 0

2 years ago

Please help I wasn’t here for this lesson

katovenus [111]

Answer:

27 cm squared

Explanation:

have a good rest of ur day

8 0

2 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

In recent years, assistive technologies have been developed to give people with disabilities equal access to resources. Apex

rosijanka [135]

Answer: as the public need for equal access for people with disabilities became understood, laws were enacted to mandate assistive technologies inclusion in almost all public spaces

Explanation: I just took the test but my teachers a füçkįñg čūńt so don't know if the this will help

3 0

3 years ago

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.

victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A) = N₀(B)

N = N₀ $e^{-\lambda t}$

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ $e^{-\lambda_1 t}$

For B

N(B) = N(B)₀ $e^{-\lambda_2 t}$

N(A) = N(B) , for t = 2 h

N(A)₀ $e^{-\lambda_1 t}$ = N(B)₀ $e^{-\lambda_2 t}$

N(A)₀ $e^{-\lambda_1 t}$ = 5 x N₀(A) $e^{-\lambda_2 t}$

$e^{-\lambda_1 t}$ = 5 $e^{-\lambda_2 t}$

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

half life = .693 / λ

For A

.77 = .693 / λ₁

λ₁ = .9 h⁻¹

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

Putting t = 2 h , λ₁ = .9 h⁻¹

$e^{\lambda_2\times 2}$ = 5 $e^{.9\times 2}$

$e^{\lambda_2\times 2}$ = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0

2 years ago