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miv72 [106K]
3 years ago
13

What is the fastest plane in the world

Engineering
2 answers:
Aleonysh [2.5K]3 years ago
7 0

Answer:

Lockheed SR-71 Blackbird

Explanation:

Phoenix [80]3 years ago
5 0
Lockheed SR-71 Blackbird
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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
The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi
leva [86]

Answer:

The distance measure from the wall = 36ft

Explanation:

Given Data:

w = 10

g =32.2ft/s²

x = 2

Using the principle of work and energy,

T₁ +∑U₁-₂ = T₂

0 + 1/2kx² -wh = 1/2 w/g V²

Substituting, we have

0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²

170 = 0.15528V²

V² = 170/0.15528

V²     = 1094.796

V = √1094.796

V = 33.09 ft/s

But tan ∅ = 3/4

∅ = tan⁻¹3/4

   = 36.87°

From uniform acceleration,

S = S₀ + ut + 1/2gt²

It can be written as

S = S₀ + Vsin∅*t + 1/2gt²

Substituting, we have

0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)

19.85t - 16.1t² + 3 = 0

16.1t² - 19.85t - 3 = 0

Solving it quadratically, we obtain t = 1.36s

The distance measure from the wall is given by the formula

d = VCos∅*t

Substituting, we have

d = 33.09 * cos 36. 87 * 1.36

d = 36ft

5 0
3 years ago
Read 2 more answers
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