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3 years ago

10

A lighthouse built at sea level is 170ft high from its top , the angle of depression of a buoy is 29 degrees . Find the distance

from the buoy to the foot of the lighthouse

1 answer:

aivan3 [116]3 years ago

6 0

Answer: c

Explanation:

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A microwave transmitter has an output of 0.1W at 2 GHz. Assume that this transmitter is used in a microwave communication system

Len [333]

Answer:

gain = 353.3616

P_r = 1.742*10^-8 W

Explanation:

Given:

- The output Power P_o = 0.1 W

- The diameter of the antennas d = 1.2 m

- The frequency of signal f = 2 GHz

Find:

a. What is the gain of each antenna?

b. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the available signal power out of the receiving antenna.

Solution:

- The gain of the parabolic antenna is given by the following formula:

gain = 0.56 * 4 * pi^2 * r^2 / λ^2

Where, λ : The wavelength of signal

r: Radius of antenna = d / 2 = 1.2 / 2 = 0.6 m

- The wavelength can be determined by:

λ = c / f

λ = (3*10^8) / (2*10^9)

λ = 0.15 m

- Plug in the values in the gain formula:

gain = 0.56 * 4 * pi^2 * 0.6^2 / 0.15^2

gain = 353.3616

- The available signal power out from the receiving antenna is:

P_r = (gain^2 * λ^2 * W) / (16*pi^2 * 10^2 * 10^6)

P_r = (353.36^2 * 0.15^2 * 0.1) / (16*pi^2 * 10^2 * 10^6)

P_r = 1.742*10^-8 W

4 0

3 years ago

Advocard [28]

Answer:

I would say C

Explanation:

let me know if im right

3 0

1 year ago

Read 2 more answers

A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at on

MAXImum [283]

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

$f_n=\frac{nv}{2L}$

For (n+1)th value the frequency is

$f_{n+1}=\frac{(n+1)v}{2L}$

Taking a ratio of both equation and solving for n such that the value of n is a whole number

$\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\$

So n is a whole number this means that the pipe is open ended.

For confirmation the nth frequency for a closed ended pipe is given as

$f_n=\frac{(2n+1)v}{4L}$

For (n+1)th value the frequency is

$f_{n+1}=\frac{(2n+3)v}{4L}$

Taking a ratio of both equation and solving for n such that the value of n is a whole number

$\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\$

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

$f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m$

The value of length is 3.4m.

5 0

3 years ago

Systematic searching is a skill that takes ________ to master.

bagirrra123 [75]

Answer: B, repetitive practice! hope this helps. :)

Explanation:

7 0

3 years ago

Read 2 more answers

A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic el

elena55 [62]

Answer:

E= 15 GPa.

Explanation:

Given that

Length ,L = 0.5 m

Tensile stress ,σ = 10.2 MPa

Elongation ,ΔL = 0.34 mm

lets take young modulus = E

We know that strain ε given as

$\varepsilon =\dfrac{\Delta L}{L}$

$\varepsilon =\dfrac{0.34}{0.5\times 1000}$

$\varepsilon =0.00068$

We know that

$\sigma = \varepsilon E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa$

Therefore the young's modulus will be 15 GPa.

8 0

3 years ago