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ira [324]
3 years ago
11

AM radio signals have frequencies between 550 kHz and 1600 kHz (kilohertz) and travel with a speed of 3.0Ã108m/s. What are the w

avelengths of these signals?
Physics
1 answer:
Westkost [7]3 years ago
5 0

Answer:

The wavelength of these signals is as follow:

  • Wavelength of 550 kHz is 545.45 m
  • Wavelength of 1600 kHz is 187.5 m

Explanation:

Given that:

Frequency = 550 kHz & 1600 kHz

Velocity = 3.0 x 10⁸ m/s

As we know that frequency is expressed by the following equation:

  • Frequency = Velocity / Wavelength ---- (1)

For 550 kHz:

The equation can be rearranged as

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (550 x 1000 Hz)

Wavelength = 545.45 m

For 1600 kHz:

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (1600 x 1000 Hz)

Wavelength = 187.5 m

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Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0
3 years ago
Can I put an “I” for an argumentative essay?
masha68 [24]

Answer:

There is no place for first person in formal academic argumentative essay. However, you can use first person only if it has been asked to provide your personal opinion or reflection about the topic.

Explanation:

4 0
3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
A plane flew for 2 hours at 467 mph and 5 hours at 536 mph how far did the plane fly in miles?
marysya [2.9K]

Answer:

3614

Explanation:

3 0
3 years ago
A carbon resistor is 4 mm long and has a constant cross section of 0.2 mm2. The conductivity of carbon at room temperature is 3
tangare [24]

Answer:

The resistance is found to be 6Ω

The current is found to be 0.66 A

Explanation:

The resistance of a conductor in terms of its dimensions is given as:

R = ρL/A

where,

R = resistance = ?

ρ = resistivity = 3 x 10⁴ Ω.m

L = Length = 4 mm = 0.004 m

A = Cross-sectional area = 0.2 mm² = 0.2 x 10⁻⁶ m²

Therefore,

R = (3 x 10⁴ Ω.m)(0.004 m)/(0.2 x 10⁻⁶ m²)

<u>R = 6 Ω</u>

Now, the potential difference between both ends of the resistor is:

ΔV = 16 V - 12 V = 4 V

Now, from Ohm's Law:

V = IR

I = V/R

I = 4 V/ 6 Ω

<u>I = 0.66 A</u>

7 0
3 years ago
Read 2 more answers
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