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Mandarinka [93]
3 years ago
13

List the following types of electromagnetic radiation in order of increasing wavelength:(i) the gamma rays produced by a radioac

tive nuclide used in medical imaging(ii) radiation from an FM radio station at 93.1 MHz on the dial(iii) a radio signal from an AM radio station at 680 kHz on the dial(iv) the yellow light from sodium-vapor streetlights(v) the red light of a light-emitting diode, such as in a calculator display
Physics
1 answer:
Naddik [55]3 years ago
3 0

Answer:

In order of increasing wavelength, the answer is:

(i) The gamma rays produced by a radioactive nuclide used in medical imaging

(iv) The yellow light from sodium-vapor streetlights

(v) The red light of a light emitting diode, such as in a calculator display

(ii) Radiation from an FM radio station at 93.1 MHz on the dial

(iii) A radio signal from an AM radio station at 680kHz on the dial

Explanation:

First, you have to know that the wavelength of a sinusoidal wave traveling at a constant speed is given by:  

                                       λ = \frac{v}{f}

Where λ is the wavelength, v is the constant speed and f is the wave's frequency. In the case of electromagnetic radiation in free space, the constant speed is the speed of light.

From explained above, you can conclude that there is a proportionality relationship between the wavelength and the frequency, they are inversely proportional. That means: the highest frequency will have the shortest wavelength and vice-versa.

So, you have the following types:

(i) The gamma rays produced by a radioactive nuclide used in medical imaging

Frequency : Typically greater than 10^{19} Hz

(ii) Radiation from an FM radio station at 93.1 MHz on the dial

Frequency: 93.1 MHz

(iii) A radio signal from an AM radio station at 680 kHz on the dial

Frequency: 680 kHz

(iv) The yellow light from sodium-vapor streetlights

Frequency: Visible spectrum of approx. 508 - 526 THz

(v) The red light of a light-emitting diode, such as in a calculator display

Frequency: Visible spectrum of approx. 400 - 484 THz

Then, you have to organize them from the highest frequency to the smallest one (decreasing frequency), and as the highest frequency will have the shortest wavelength, you are going to have it organized in an increasing wavelength mode.

Then in order of increasing wavelength, the answer will be:

                                           (i) , (iv), (v), (ii), (iii)

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Answer:

Explanation:

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= \frac{\mu_0}{4\pi} \frac{2i}{r}

i is current , r is distance from wire

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force on second wire per unit length

B I L , I is current in second wire , L is length of wire

=  10⁻⁷ x \frac{2\times49}{r} x 33 x 1

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This should balance weight of second wire per unit length

3234 x \frac{10^{-7}}{r} = .075

r = \frac{3234}{.075} x 10⁻⁷

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3 years ago
Just need help with 1 and 2 please :D i’m having a bit of trouble :/
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Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want t
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Answer:

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The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

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The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

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r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

Hope this Helps!!!

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Answer:

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3 years ago
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