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tatiyna
3 years ago
6

Your EMS system covers a large area. For reliable transmission between mobile and portable​ radios, which of the following is​ e

ssential?
A. Cell phonesB. Microwave radiosC. RepeatersD. Digital radios
Physics
1 answer:
Effectus [21]3 years ago
5 0

Answer:

Repeaters

Explanation:

EMS are communication systems that coordinates emergency medical care among ambulances, 911 (telephone) dispatch centers, and hospital emergency departments. According to my research on EMS systems, I can say that based on the information provided within the question you would need Repeaters. These are a piece of hardware that repeats and extends a signal so that it reaches a greater distance.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

y = \frac{m\lambda}{d}

Here,

m = Number of order bright fringe

\lambda = Wavelength

d = Distance between slits

Both distance are the same, then

y_1 = y_2

\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}

\frac{m_1\lambda_1}{m_2\lambda_2} =1

 Rearranging to find the second wavelength

m_1 \lambda_1 = m_2 \lambda _2

\lambda_2 = \frac{m_1\lambda_1}{m_2}

\lambda_2 = \frac{7(629)}{8}

\lambda_2 = 550.3nm

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

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2 years ago
Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in
Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

L = \frac{N^2 \mu_0 A}{l}

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

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