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Svetllana [295]
1 year ago
12

Explain how smart materials can be used by manufacturers to improve health and safety for children's products and goods.​

Engineering
1 answer:
Ierofanga [76]1 year ago
6 0

...simplify devices, reducing weight and the chance of failure.

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. Chemical manufacturers must present which Information on the product's label?
STatiana [176]

Answer: A) Product identifier

Explanation:

A product identifier is a means, name or number used to identify hazardous and dangerous chemicals on a label or in the SDS( safety data sheet). It provides a special means which helps the user to identify the chemical. Any product identifier used should permit a cross-references to be made among the listed hazardous chemicals required in the written hazard communication program, and the label including SDS(safety data sheet).

5 0
3 years ago
2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m
Daniel [21]

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

4 0
3 years ago
As temperature decreases a batteries availability power what
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The answer is increases because when something like that decreases it’s always decreasing that probly makes no sense Imao but it’s increases
7 0
3 years ago
1. You use
lorasvet [3.4K]
4-ways tell me if I’m wrong
4 0
3 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
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