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SpyIntel [72]
3 years ago
7

During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz elec

tromagnetic wave. (a) What is the maximum electric field strength created? (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? (c) What is the wavelength of the electromagnetic wave?
Physics
1 answer:
erik [133]3 years ago
4 0

Answer:

(a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is 4.433\times10^{-11}\ T

(c). The wavelength of the electromagnetic wave is 3\times10^{8}\ m

Explanation:

Given that,

Maximum potential = 4.00 mV

Distance = 0.300\ m

Frequency = 1.00 Hz

(a). We need to calculate the maximum electric field strength

Using formula of the potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

E=\dfrac{4.00\times10^{-3}}{0.300}

E=0.0133\ V/m

(b). We need to calculate the maximum magnetic field strength in the electromagnetic wave

Using formula of the maximum magnetic field strength in the electromagnetic wave

B=\dfrac{E}{c}

Put the value into the formula

B=\dfrac{0.0133}{3\times10^{8}}

B=4.433\times10^{-11}\ T

(c). We need to calculate the wavelength of the electromagnetic wave

Using formula of wavelength

c=f\lambda

\lambda=\dfrac{c}{f}

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{1.00}

\lambda=3\times10^{8}\ m

Hence, (a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is 4.433\times10^{-11}\ T

(c). The wavelength of the electromagnetic wave is 3\times10^{8}\ m

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where <em>m</em> = 12.5 kg and <em>g</em> = 9.80 m/s².

The minimum force <em>F</em> needed to overcome <u>maximum</u> static friction <em>f</em> and get the object moving is

<em>F</em> > <em>f</em> = 0.50 <em>n</em> = 61.25 N ≈ 61.3 N

which means a push of <em>F</em> = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.

So:

(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.

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MODERN PHYSICS
frosja888 [35]

Answer:

D. n=6 to n=2

Explanation:

Given;

energy of emitted photon, E = 3.02 electron volts

The energy levels of a Hydrogen atom is given as;  E = -E₀ /n²

where;

E₀ is the energy level of an electron in ground state =  -13.6 eV

n is the energy level

From the equation above make n, the subject of the formula;

n² = -E₀ / E

n² = 13.6 eV / 3.02 eV

n² = 4.5

n = √4.5

n = 2

When electron moves from higher energy level to a lower energy level it emits photons;

E = E_0(\frac{1}{n_1^2}-\frac{1}{n_2^2} )\\\\\frac{1}{n_1^2}-\frac{1}{n_2^2} = \frac{E}{E_o} \\\\\frac{1}{4} -\frac{1}{n_2^2} = \frac{3.02}{13.6} \\\\\frac{1}{4} -\frac{1}{n_2^2} =0.222\\\\\frac{1}{n_2^2} = 0.25 - 0.22\\\\\frac{1}{n_2^2} = 0.03\\\\n_2^2 = 33.33\\\\n_2 = \sqrt{33.33} = 6

For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2

Therefore,  The electron energy-level transition is from n = 6 to n = 2

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3 years ago
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