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3 years ago

Radio waves travel at the speed of light. What is the wavelength of a radio signal with a frequency of 9.45 x 10^7 Hz?

Physics

1 answer:

Klio2033 [76]3 years ago

3 0

Answer:

So I never really knew you

God, I really tried to

Blindsided, addicted

Felt we could really do this

But really I was foolish

Hindsight, it's obvious

Explanation:

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A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

5 0

3 years ago

An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 7 m/s over

stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

= 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .

xf =xi+vi*t+1/2*a*t

t=√2(xf-xi)/a

t=3.765×10^-10 s

now we can use the power P Eq.

P=W/Δt => ΔK/Δt

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0

3 years ago

When a person speaks, a sound intensity is generated that is 600 times greater than when the person whispers. What is the differ

Charra [1.4K]

Answer:

Originally : Level = log I / I0

Currently: Level = 10 log I / I0

Level = 10 log 600 = 10 * 2.78 = 27.8

Note the term 1 bel = 10 decibels

5 0

2 years ago

A car travels 40 miles in 30 minutes.

lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$

=>35.75 m/s----------------------------------------------------------(1)

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

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

$S= ut+\frac{1}{2}at^2$

Substituting the known values,

$s= ut+\frac{1}{2}at^2$

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

a= - 2.38 $m/s^2$ ----------------------------------------(4)

Now substituting 4 in 3 we get

$s=566.25+\frac{(225( - 2.38)}{2}$

$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

s=268.8m--------------------------------------------------------------(5)

4 0

3 years ago

Joe has a mass of 40 kg, and Sally has a mass of 25 kg. They sit in class with a separation of 6 meters. Find the gravitational

goldfiish [28.3K]

Answer:

hmmmmmmmm

Explanation:

mmmmmmmmmmmmmmmmmmmmmmm pay attention in class kid

4 0

3 years ago