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amm1812
3 years ago
7

In unguided medium (free space), the electromagnetic (EM) signal wave spreads as it leaves the transmit antenna. Since the power

of the EM signal resides at the surface area of the wave front, signal power is described as signal power density (i.e., watts per area).
a. True
b. False
Engineering
1 answer:
Daniel [21]3 years ago
5 0

Answer:

a. True

Explanation:

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What is the basic formula for actual mechanical advantage?
slavikrds [6]

Answer:

Mechanical Advantage Formula

The efficiency of a machine is equal to the ratio of its output to its input. It is also equal to the ratio of the actual and theoretical MAs. But, it does not mean that low-efficiency machines are of limited use. An automobile jack, for example, have to overcome a great deal of friction and therefore it has low efficiency. But still, it is extremely valuable because small effort can be applied to lift a great weight.

Also, in another way the mechanical advantage is the force generated by a machine to the force applied to it which is applied in assessing the performance of the machine.

The mechanical advantage formula is:

MA = FBFA

Explanation:

MAmechanical advantageFBthe force of the object

FAthe effort to overcome the force

3 0
3 years ago
You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the
defon
You need to explain it more simple as everyone is clueless
7 0
2 years ago
A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b
Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= \frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= \frac{8640*100-8640*20}{20*10^5}

k= 0.35 per day

6 0
3 years ago
A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j
Anna [14]

Answer:

the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

Explanation:

Given that;

volume of cut = 25,100 m³

Volume of dry soil fill = 23,300 m³

Weight of the soil will be;

⇒ 93% × 18.3 kN/m³ × 23,300 m³

= 0.93 × 426390 kN 3

= 396,542.7 kN  

Optimum moisture content = 12.9 %

Required amount of moisture = (12.9 - 8.3)% = 4.6 %

So,

Weight of water required = 4.6% × 396,542.7 = 18241 kN

Volume of water required = 18241 / 9.81 = 1859 m³

Volume of water required = 1859 kL

Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

6 0
3 years ago
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