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Anna71 [15]
3 years ago
15

If you are driving a 30-foot

Engineering
1 answer:
Ratling [72]3 years ago
5 0

Answer:

3 sec

If you are driving a 30-foot vehicle at 55 mph, how many seconds of following distance should you allow? 30ft truck. = 3 sec. Since the truck is over 40 mph.

Explanation:

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What should you consider when choosing the type of head protection
Serggg [28]

Answer: hope this helps u

Explanation:

Material,Flexibility,Comfort,Workplace Hazards

6 0
2 years ago
What type of degree is needed to be an architect?
Rina8888 [55]
Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license.
3 0
3 years ago
Read 2 more answers
Nonshielded cable with a 1.5-inch diameter should have a minimum bending radius of
Flura [38]

Answer:

c is the answer because we have to double the number

6 0
2 years ago
4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and
Alika [10]

Answer:

The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s

Peak flow of the aggregated runoff hydrograph is 420.58 m³/s

The total volume of runoff is 2125000 m³/s

Explanation:

We have

A = 25 km²

tr = 10 min = 1/6 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr

Qp = 2.08×25/1.043 = 49.84 m³/s

Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr

 

Since the area is  

Time (min)           Runoff (cm)       Volume of runoff m³

0                   0                                     0

10                  4                                     1000000 m³

20                 2.5                                  625000 m³

30                 2                                      500000 m³

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

For the 1st  10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×4/1.043 = 197.92 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

 

For the 2nd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 123.7 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

For the 3rd 10 minutes, we have

A = 25 km²

tr = 30 min = 1/2 hr

tc = 1.6 hr

lag time = 0.6 tc = 0.96 hr

Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr

Qp = 2.08×25×2.5/1.043 = 98.96 m³/s

Tb = 8/3×Tp = 8/3×1.21 = 3.227  hr

 

Peak flow of aggregate runoff is given by

Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s

Total volume of runoff is given by

Total volume of runoff = 1000000 + 625000 + 500000 =  2125000 m³/s

6 0
3 years ago
We are used to connect the external devices and power supply to a computer. My anagram is torps
lutik1710 [3]

Answer: Ports  

Explanation:

Ports are the devices tat is used for connection of input devices and output devices wit computer.Te links created by ports between peripheral and computing system connects devices such as monitors, keyboard,modems , printer speaker s etc.  

External devices get plugged into ports which are embedded on motherboard of the computing system to establish peripheral connection.

4 0
3 years ago
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