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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>

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A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time

baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

$i = \frac{q}{t}$

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

$i = increased$

So correct answer will be

B. The current increases.

4 0

3 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$

Here,

n = Number of electrons

e = Charge of each electron

$n = \frac{q}{e}$

Replacing,

$n = \frac{3.002*10^{-12}}{1.6*10^{-19}}$

$n = 2.44*10^7$

Therefore the number of electrons that reside on the drop is $2.44*10^7$

5 0

3 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

3 years ago

Order from biggest to smallest?<br> Milky Way, universe, planets, clusters, and stars

Zinaida [17]

Hello,

The answer is "universe, Milky Way, clusters, stars, planets".

Reason:

The universe would be the biggest because it has all the galaxy's, starts, clusters, and planets into one. Then it would be Milky Way because this is a galaxy that contains: stars, planets, and clusters. Then it would be clusters because that contains stars, or planets in one group. Then be stars because stars are bigger than planets. Then it would be planets. Therefore the order should go like this: <span>Milky Way, universe, planets, clusters, and stars.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>

5 0

3 years ago

Read 2 more answers

A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What

Jobisdone [24]

Answer:

$v_{avg} = 6.41 m/s$

Explanation:

Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

here we know that initially the sky diver drops without opening parachute by total displacement 625 m

then she open her parachute and drop another 362 m

so first it took time t = 15 s to drop without open parachute

then it took t = 139 s to drop next displacement

so here total displacement is given as

$d = 625 m + 362 m$

total time is given as

$t = 15 s + 139 s$

so average velocity is given as

$v_{avg} = \frac{625 + 362}{15 + 139}$

$v_{avg} = 6.41 m/s$

4 0

3 years ago