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daser333 [38]
3 years ago
11

(Practice work, not graded)

Engineering
1 answer:
luda_lava [24]3 years ago
7 0

Answer: ALL CAREFULLY ANSWERED CORRECTLY.

Explanation:

1) A loaf of Bread PHYSICAL SYSTEM

✓ How can the environment affect the edibility of the bread

✓ What are the constituents that makes up the bread

✓ What process is involved in these constituents mixing to form the loaf.

2) The law of thermodynamics makes us to understand that when heat/energy passes through a system, the systems internal energy changes with respect to the conservation of energy law. That is energy lost = energy gained. Typically, ice would melt in a cup of hot tea because of the thermal energy in the molecules of the hot tea. When you heat a material, you are adding thermal kinetic energy to its molecules and usually raising its temperature. The temperature of the ice raises due to the kinetic energy added to it and it melts to water.

3) The theory of systems view the world as a complex system of interconnected parts. If we consider the society; (financial systems, political systems, etc) we will agree that they individually have their own components and it's the summation of this components that makes the system, this implies that system thinking could be applicable in this kinda of systems as long as they are made up of components.

4) Technology has boosted every sector of our lives and it has the capacity to do more. Restricting it's importance to entertainment alone would be an underusing of its potentials. Engineering students infact should not need any drive to be encouraged about maximizing all it can do in shaping our world.

5) ~ Nature shows its splendid soul

~Never ceases to leave us in amazement

~And we are in love

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A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th
Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0
3 years ago
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