Ask question

Ask question

All categories

3 years ago

10

An fm radio station broadcasts at a frequency of 97 mhz . what inductance should be paired with a 7.0 pf capacitor to build a re

ceiver circuit for this station?

1 answer:

SIZIF [17.4K]3 years ago

8 0

<span>xL = xC </span>

<span>ωL = 1/ωC </span>

<span>L = 1/(ω²C) </span>

<span>L = 1/((2πf)²C) </span>

<span>L = 1/((2π * 91 E6)²(10 E-12)) </span>

<span>L = 3.059 E-7 H </span>

You might be interested in

What do you think will happen to dirt and rocks on the Mountainside when the ice melts

AlekseyPX

This question is personal preference... just answer with whatever YOU think, there probably isn’t s wrong answer if I’d guess.

3 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

$\Delta U = q \Delta V$

where

q is the charge transferred

$\Delta V$ is the potential difference between the ground and the clouds

Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

$K=\frac{1}{2}mv^2$

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

5 0

3 years ago

A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca

Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, $f_o=30\ cm$

The focal length of the eyepiece lens of an astronomical telescope, $f_e=5\ mm=0.5\ cm$

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

$m=\dfrac{-f_o}{f_e}$

$m=\dfrac{-30}{0.5}$

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

7 0

3 years ago

What temperature is Sirius B?

Dmitry [639]

The answer would be 9,940 K is the temperature of Sirius B.

5 0

3 years ago

A single conservative force acts on a 5.30-kg particle within a system due to its interaction with the rest of the system. The e

cupoosta [38]

Answer:

Given that

m = 5.3 kg

Fx = 2x + 4

We know that work done by force F given as

w= ∫ F. dx

a)

Given that x=1.08 m to x=6.5 m

Fx = 2x + 4

w= ∫ F. dx

$w=\int_{1.08}^{6.5}(2x+4) .dx$

$w=\left [x^2+4x \right ]_{1.08}^{6.5}$

$w=(6.5^2-1.08^2)+4(6.5-1.08)\ J$

w=62.7 J

b)

We know that potential energy given as

$F=-\dfrac{dU}{dx}$

∫ dU = -∫F.dx ( w= ∫ F. dx)

ΔU= -62.7 J

c)

We know that form work power energy theorem

Net work = Change in kinetic energy

W= KE₂ - KE₁

62.7 =KE₂ - (1/2)x 5.3 x 3²

KE₂ = 86.55 J

This is the kinetic energy at 6.5m

8 0

3 years ago