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shepuryov [24]
3 years ago
6

The AGC control voltage: ___________

Engineering
1 answer:
lyudmila [28]3 years ago
3 0

Answer:

The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.

Explanation:

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What type of footwear protects your toes from falling objects and being crushed?
Lady bird [3.3K]

Answer:Steel toe

Explanation:

6 0
4 years ago
Read 2 more answers
A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac
WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

     = 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2

     =70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

    = (73.17 -53.3/53.3)*100

    = 37%

5 0
3 years ago
What financial arguments could you use to justify your proposed
Gnoma [55]

The recommendation to segregate FLTs and the workers are as follows:-1)Reputation of warehouse:- To be in the market the reputation of warehouse should be good,it can only happen when the worker of that warehouse is happy with the management looks after worker external and internal affairs. There should be two pathways one for vehicle and other for walking in which both can’t use vice versa.2)High Profitability:- When there is no incident or accident happens between the FLTs and the workers in the warehouse then off course the worker will be regular at work then there will be high profit .3)Insurance premium:- If there is zero accident happens in the ware house then there will no claim, the company will be in the profit..

7 0
3 years ago
A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g
SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

3 0
3 years ago
An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s
nignag [31]

Given:

X_{L} = 50.0 \ohm

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

V_rms} = 85.0 V

Solution:

To calculate max current in inductor, I_{L(max):

At f = 60.0 Hz

X_{L} = 2\pi fL

50.0 = 2\pi\times 60.0\times L

L = 0.1326 H

Now, reactance X_{L} at f' = 45.0 Hz:

X'_{L} = 2\pi f'L

X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm

Now, I_{L(max) is given by:

I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}

I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A

Therefore,  max current in the inductor, I_{L(max) = 2.13 A

7 0
3 years ago
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