A radio station has a frequency of 90.9 megahertz (9.09 × 10^7 Hz). What is the wavelength of the radiowaves the station emits from its radio tower?

Answers

Answer 1
Answer: well it depends of the distance, but u get your frequency and u times it by a round number if im correct
Answer 2
Answer:

Answer:

well it depends of the distance, but u get your frequency and u times it by a round number if im correct

Explanation:


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The copper wire and bulb is connected in a series with 220 V electric supply. Why only an electric bulb glows where as the copper wire remains the same.Give REASON to support the answer.

Answers

-- In a series circuit, the current ( I ) is the same at every point.

-- The power dissipated by any section of the circuit is I² x Resistance.

-- The wire has very low resistance, so I²R is very low dissipated power.

-- The filament in the bulb has most all of the resistance in the circuit,
so it dissipates virtually all the power of the circuit, and certainly much
more than the wires do.

Voltage sources are used to sustain a. the flow of protons
b. chemical reactions
c. resistance
d. the movement of electric charge

Answers

Answer:

d. the movement of electric charge

Explanation:

Voltage source is used to provide constant voltage difference across the ends of a given conductor where it is connected

Now when voltage source is connected across the conductor we can say that due to constant voltage difference there would be a constant current will maintain in the conductor

It is given by

i = (\Delta V)/(R)

now since the voltage difference and resistance both are constant so current will be same in the circuit

So here the flow of charge will remain constant due to voltage source

So correct answer would be

d. the movement of electric charge

the movement of electric charge

A truck accelerates 2 m/s2 when it is empty. If the truck is filled so that it has twice the mass and the same amount of net force is applied, how much will the truck accelerate?8 m/s2
1 m/s2
4 m/s2
2 m/s2

Answers

Answer :

a = 1 m/s^2

Explanation:

Given information

Net force is a constant

The mass of the truck is m_(o)

Initial acceleration a_(o)= 2 m/s^(2)

The mass of the truck is increased twice as much

F_(net)= m_(o) a_(o) = ma

F_(net)= m_(o) a_(o) = 2m_(o)a

a=\frac{a_(o) } {2}

a = 2/2

a = 1 m/s^(2)

This is a conceptual problem, I’ll try to upload a picture:

A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.43 V and a current of 3.1 A are induced in the coil. The wire is then re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What (a) emf and (b) current are induced in the square coil

Answers

Answer:

2.62A

Explanation:

Given

V = 0.43 V

I = 3.1 A

Then, V = IR, R = V/I

R = 0.43/3.1

R = 0.14 Ω

The induced emf = dB/dt * A

So that, dB/dt = emf/A

Since dB/dt is constant then Emf/A(circle) = Emf/A square

So Emf (square)/Emf (circle) = A square / A circle

A circle = πr². The perimeter of the square is 2πr which also is the circumference of the square.

Since the perimeter is 2πr, then each side would be πr/2. Thus, the area of the square would be, (πr/2)² = π²r²/4

So A square/Acircle = (π²r²/4) / πr² = π/4 = 0.79

this means that, emf square = emf circle * 0.79

emf square = 0.43*0.79 = 0.34V

I = V/R

I = 0.34/0.13

I = 2.62A

ListenA projectile with mass m is fired with initial horizontal velocity vx from height h above level ground. Which change would have resulted in a greater time of flight for the projectile? [Neglect friction.]decreasing the mass to m/2decreasing the height to h/2increasing the initial horizontal velocity to 2vxincreasing the height to 2h

Answers

The only way to increase the time of flight is to increase the height traveled by the projectile to 2h.

The given parameters;

  • mass of the projectile, = m
  • initial horizontal velocity, = v_x

The time of flight of the projectile is calculated as follows;

h = v_yt + (1)/(2) gt^2

where;

v_y is the initial vertical velocity of the projectile = 0

h = (1)/(2) gt^2\n\ngt^2 = 2h\n\nt = \sqrt{(2h)/(g) } \n\n

  • If friction due to air resistance is neglected, the time of flight of the projectile is not affected by its mass.
  • The initial horizontal velocity is not considered in the vertical direction.

Thus, the only way to increase the time of flight is to increase the height traveled by the projectile to 2h.

Learn more here:brainly.com/question/20427663

Answer:

d)

Explanation:

a) decreasing the mass to m/2. FALSE. Mass doesn't influence the equations of accelerated motion.

b) decreasing the height to h/2. FALSE. From a lower initial height the projectile will reach the ground faster.

c) increasing the initial horizontal velocity to 2vx. FALSE. The horizontal and vertical components are independent, and time of flight is dependent only on vertical conditions in this case.

d) increasing the height to 2h. TRUE. From a higher initial height the projectile will reach the ground later.

What is the kinetic energy of a 30 kg falling object when the object reaches a velocity of 20 m/s?

Answers

Ek = 1/2mv^2 Ek = 1/2 * 30kg * 20^2 m/s Ek = 6000J Hope this Helps!