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astraxan [27]
3 years ago
8

Who does the narrator blame for the loss of her job as editor-in-chief? see if i care readworks

Engineering
2 answers:
lora16 [44]3 years ago
6 0

Answer:

idkkk

Explanation:

lora16 [44]3 years ago
3 0

Answer:

The Narrator blamed Maxwells for the loss of her job.

Explanation:

hope im not late

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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
Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes th
Svet_ta [14]

Answer:

The time required to elute the two species is 53.3727 min

Explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621

The mathematical expression for the time to elute the two species is:

\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}

Here

ReB = 1.5

t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min

6 0
3 years ago
The slope distance and zenith angle between points A and B were measured with a total station instrument as 17685.20 ft and 93°
dimaraw [331]

True

Explanation:

three point bending is better than tensile for evaluating the strength of ceramics. it is got a positive benefit to tensile for evaluating the strength of ceramics.

4 0
3 years ago
A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight
jek_recluse [69]

Answer:

a) the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb

Explanation:

Given the data in the question;

Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in

Diameter d = 3/16 in

Unit weight w = 490 pcf

First we determine the area of the wire;

A = π/4 × d²

we substitute

A = π/4 × (3/16)²

A = 0.0276 in²

Next we get the Volume

V = Area × Length of wire

we substitute

V = 0.0276 × 4800

V = 132.48 in³

Weight of the steel wire will be;

W = Unit weight × Volume

we substitute

W = 490 × ( 132.48 / 12³ )

W = 490 × 0.076666

W = 37.57 lb

a) the maximum tensile stress due to the weight of the wire;

σ_w = W / A

we substitute

σ_w = 37.57 / 0.0276

= 1361.23 psi

Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi

Maximum load P that the wire can safely support its lower end will be;

P = ( σ_{all - σ_w )A

we substitute

P = ( 24000 - 1361.23  )0.0276

P = 22638.77 × 0.0276

P = 624.83 lb

Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb

5 0
3 years ago
Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
aev [14]

Answer: 220

Explanation:

7 0
3 years ago
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