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Tom [10]
3 years ago
7

Sirens and flashing lights do NOT indicate that _____ may be nearby.

Engineering
2 answers:
Naya [18.7K]3 years ago
3 0

Answer:

A safe enveroment.

Explanation:

Sirens usally mean fire or someone died.

kaheart [24]3 years ago
3 0
Sirens or flashing lights do not indicate that something is safe is nearby it means that there are no emergency’s
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Differentiate between "Threshold and Resolution" with suitable examples.
9966 [12]

Answer:

to make the bace of a building more sturdy

Explanation:

example: the bace of the empire state building is stone very sturdy

6 0
3 years ago
Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
3 years ago
¿Que piensas respecto al hecho de que María y Efraín rara vez se comunican directamente?
AleksandrR [38]

Answer:

Crean un código usando la naturaleza.

8 0
3 years ago
Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.
atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage v_p=120volt

(i) We know that peak voltage is give by v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt

(ii) We know that for transformer \frac{v_p}{v_s}=\frac{n_p}{n_s}

So \frac{169.08}{v_s}=\frac{10}{1}

v_s=16.90volt

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt

(v) Now dc current is given by i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A

4 0
3 years ago
List, in ascending order, the cutoff frequencies for the first ten modes of a rectangular waveguide, normalized to the cutoff fr
Alisiya [41]

Answer:

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

Explanation:

Assuming a = 2b

Attached below is the required steps to the solution

The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order  is :

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

5 0
3 years ago
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