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kap26 [50]
3 years ago
10

The waves from a radio station can reach a home receiver by two paths. One is a straight-line path from transmitter to home, a d

istance of 32.0 km. The second is by reflection from the ionosphere (a layer of ionized air molecules high in the atmosphere). Assume this reflection takes place at a point midway between receiver and transmitter, the wavelength broadcast by the radio station is 344 m, and no phase change occurs on reflection. Find the minimum height of the ionospheric layer that could produce destructive interference between the direct and reflected beams.
Physics
1 answer:
slava [35]3 years ago
8 0

Answer:

Explanation:

The path of waves reaching directly and through reflection from ionosphere will form a isosceles triangle.

Let the height be h . This will be height of a triangle of equal side whose base is of length 32 km. we shall calculate the length of one side of this triangle .

this length be l

l² = 16² + h²

l =  √(256 +h² )

2l = 2√(256 +h² )

path difference = 2l - 32 km

For destructive interference

path difference = wavelength /2 for minimum height .

2l - 32  = .344/2

2√(256 +h² )- 32 = . 172

2√(256 +h² ) = 32.172

4(256 +h² ) = 1035

(256 +h² ) = 258.76

h² = 2.76

h = 1.66 km

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If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of
olga_2 [115]

Answer:

The magnitude of the electric field is 0.1108 N/C

Explanation:

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length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201}  \\\\E = 0.1108 \ N/C

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6 0
3 years ago
Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani
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Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

n = \frac{t}{t_{1/2}}

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

t_{1/2} = half-life = 700 million years

Therefore,

n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3

Now, we will calculate the number of uranium nuclei left (n_u):

n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u =  800\ million

and the rest of the uranium nuclei will become thorium nuclei (u_{th})

n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million

dividing both:

\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

6 0
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