All categories

3 years ago

- Scrap tire management is primarily regulated at the

Engineering

2 answers:

krek1111 [17]3 years ago

5 0

Answer:D

Explanation:

Scrap tire management is primarily regulated at the state

level.

kompoz [17]3 years ago

3 0

Scrap tire management is primarily regulated at the state level.

You might be interested in

How to tie shoe please

Genrish500 [490]

Step 1: Unknot Shoelaces. Make sure that the two ends of the shoelace are completely untangled and free of knots past the tongue. ...
Step 2: Create Overhand Knot. ...
Step 3: Finish the Basic Shoe Knot. ...
Step 4: Create Double Knot. ...
Step 5: Finished!

4 0

2 years ago

Which welding processes normally use a constant current power source? (Select all that apply) PLEASE HELP

Furkat [3]

Answer:

Explanation:

GTAW AND GMAW

5 0

3 years ago

Do plastic materials have high or low ductility? Explain why.

Flura [38]

The impact behavior of plastic materials is strongly dependent upon the temperature. At high temperatures, materials are more ductile and have high impact toughness. At low temperatures, some plastics that would be ductile at room temperature become brittle.

3 0

3 years ago

Explain why a hydraulic power system would be the best choice when building a device or vehicle that requires large amounts of p

ohaa [14]

Answer:because it faster

Explanation: and when building a device or vehicle that requires large

3 0

3 years ago

Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could

VladimirAG [237]

Answer:

$4265.04\ \text{m}$

$2.38\times 10^{10}\ \text{W}$

Explanation:

PE = Energy of food = 500 cal = $500\times4184=2.092\times10^6\ \text{J}$

m = Mass of object = 50 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Potential energy of food is given by

$PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}$

Nancy could raise the weight to a maximum height of $4265.04\ \text{m}$ .

Mass of $H_2$ used per year = $25\times 10^{9}\ \text{kg/year}$

Energy of $H_2$ = $\dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}$

Power

$P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}$

The power requirement is $2.38\times 10^{10}\ \text{W}$ .

6 0

3 years ago