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Alja [10]
3 years ago
12

A segment of a roadway has a free flow speed of 45 mph and a jam density of 25 ft per vehicle. Determine the maximum flow and at

a flow rate of 1950 vph
Engineering
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

2376 vph

Explanation:

Given data

free flow speed ( Vf ) = 45 mph

Jam density = 25 ft per vehicle

flow rate = 1950 vph

first we calculate the Jam density in vehicle /mile

= 5280 / 25 = 211.2 vehicle/mile

where ; 1 mile = 5280 feet

The maximum flow can be calculated using Greenshield method

q = ( Vf * jam density ) / 4  =  ( 45 * 211.2 ) / 4

  = 2376 vph

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Technician A says that ultraviolet light is bright, and very hot. Technician B says that ear protection does not need to be used
-Dominant- [34]

Answer:

a. technician a only

Explanation:

I took that worksheet

3 0
3 years ago
An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The
Mila [183]

Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

6 0
3 years ago
1.
Levart [38]

Answer:

This is a for loop.

Explanation:

6 0
3 years ago
You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______
stich3 [128]
I believe it’s D) 20 feet
3 0
3 years ago
1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0
3 years ago
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