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irakobra [83]
4 years ago
11

if you are in Birmingham AL, and you want to use your cell phone to talk to your cousin in Houston, TX, what must occur in order

for communication to be completed?
Physics
1 answer:
lesantik [10]4 years ago
3 0

Answer:

For communication to be communicated between you and your cousin who is in Houston,Texas, when a call is made by you, a request is made to the specific phone, and the telephone tower will get the request from the mobile phone. Then a signal is sent via a transmitter underground, by this the satellite communicates with the local receiver in Houston, that is linked to the local tower over there, the tower would request for your cousin's number and connects the two of you, once a link is set.

Explanation:

From the example stated, what is required for such for a far distance, is a communication satellite link.

When a call is made by you, the a connection request is sent to the specified phone.The telephone tower receives the request from The mobile phone. The local tower(Birmingham,Al) is linked to a ground transmitter by the means of a Fiber optical cable.

A signal is sent to satellite via the ground transmitter.The satellite then set's off the local receiver in (Houston,Texas) which on it's end is connected to the local tower there. This tower then ask for your cousin's mobile for a call that will be incoming, a link is set, once he/she receives the call, from there a conversation can be done.

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Answer:

(A) 0.1842 T (B) 1.3507\times 10^{-4}J/m^3 (C) 0.2188 J (D) 5.40\times 10^{-5}H

Explanation:

Length of the solenoid L = 27  cm =0.27 m

Area of cross section A=0.6\ cm^2=0.6\times 10^{-4}m^2

Number of turns N = 440

Current i = 90 A

(A) Magnetic field in the solenoid B=\mu _0ni=\frac{\mu _0Ni}{l}=\frac{4\pi \times 10^{-7}\times 440\times 90}{.27}=0.1842\ T

(B) The energy density is given by energy\ density =\frac{B^2}{2\mu _0}=\frac{0.1842^2}{2\times 4\pi \times 10^{-7}}=1.3507\times 10^4\ j/m^3

(C) The total energy contained in the coli magnetic field = energy density ×volume = energy density ×l×A

So the total energy =1.3507\times 10^4\times 0.27\times 0.6\times 10^{-4}=0.2188\ j

(D) The energy stored in the inductor is given by U=\frac{1}{2}Li^2

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An infinite line of charge with charge density λ1 = 0.6 μC/cm is aligned with the y-axis.
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Answer:

1440 × 10⁴ N/C

Explanation:

Data provided in the question:

Charge density λ1 = 0.6 μC/cm = 6 × 10⁻⁵ C/m

a = 7.5 cm = 0.075 m

Now,

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Ex(P) = \frac{2}{2}\times\frac{\lambda}{2\pi\epsilon_0a}=\frac{2\lambda}{4\pi\epsilon_0a}

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Ex(P) = \frac{2\times6\times10^{-5}\times9\times10^9}{0.075}

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