Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Distance between point A and point B is 5.0 meters and they are at the same level. (Ignore the frictional losses)


Answer 1


0.556 Watts


w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

s=ut+(1)/(2)at^2\n\Rightarrow a=(2* (s-ut))/(t^2)\n\Rightarrow a=(2* (5-0))/(29^2)=(10)/(481)

Mass of the body


Force required to move the body

F=ma\n\Righarrow F=(762)/(9.81)* (10)/(481)

Velocity of object

v=u+at\n\Rightarrow v=0+(10)/(481)* 29\n\Rightarrow v=(10)/(29)


P=Fv\n\Rightarrow P=(762)/(9.81)* (10)/(481)* (10)/(29)=0.556\ W

∴ Amount of power required to move the object is 0.556 Watts

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The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.



Specific weight of the pop, w_(s) = 8619.45 N/m^(3)

Density of the pop, \rho_(p) = 8790.76 kg/m^(3)

g_(s) = 8.79076

w_(w) = 9782.36 N/m^(3)


Volume of pop, V = 360 mL = 0.36 L = 0.36* 10^(-3) m^(3)

Mass of a can of pop , m = 0.369 kg

Weight of an empty can, W = 0.153 N


Now, weight of a full can pop, W

W' = mg = 0.369* 9.8 = 3.616 N

Now weight of the pop in can is given by:

w = W' - W = 3.616 - 0.513 = 3.103 N


The specific weight of the pop, w_(s) = (weight of pop)/(volume of pop)

w_(s) = (3.103)/(0.36* 10^(- 3)) = 8619.45 N/m^(3)

Now, density of the pop:

\rho_(p) = (w_(s))/(g)

\rho_(p) = (86149.45)/(9.8) = 8790.76 kg/m^(3)


Specific gravity, g_(s) = (\rho_(p))/(density of water, \rho_(w))


g_(s) = (8790.76)/(1000) = 8.79076

Now, for water at 20^(\circ)c:

Specific density of water = 998.2 kg/m^(3)

Specific gravity of water = 0.998 kg/m^(3)

Specific weight of water at 20^(\circ)c:

w_(w) = \rho_(20^(\circ))* g = 998.2* 9.8 = 9782.36 N/m^(3)

Assignment 1: Structural Design of Rectangular Reinforced Concrete Beams for Bending Perform structural design of a rectangular reinforced concrete beam for bending. The beam is simply supported and has a span L=20 feet. In addition to its own weight the beam should support a superimposed dead load of 0.50 k/ft and a live load of 0.65 k/ft. Use a beam width of 12 inches. The depth of the beam should satisfy the ACI stipulations for minimum depth and be proportioned for economy. Concrete compressive strength f’c = 4,000 psi and yield stress of reinforcing bars fy = 60,000 psi. Size of stirrups should be chosen based on the size of the reinforcing bars. The beam is neither exposed to weather nor in contact with the ground, meaning it is subjected to interior exposure.
• Use the reference on "Practical Considerations for Rectangular Reinforced Concrete Beams"
• Include references to ACI code – see slides from second class
• Include references to Tables from Appendix A
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Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.


We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL


W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"


d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²


ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

When looking at a chain of processes with a low yield (high defective rate), what is a good place to start investigating the source of variation?


The most upstream process with issues would be a good location to start exploring the cause of the variance.

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A manufacturing technique would be a specific procedure for generating a commodity.

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Thus the response above is correct.

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A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?



\eta _(max) = 0.2413 = 24.13%

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T_(1max) = 100^(\circ) = 273 + 100 = 373 K

operating temperature of heat engine, T_(2) = 10^(\circ) = 273 + 10 = 283 K

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For a  reversible cycle, maximum efficiency, \eta _(max) is given by:

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\eta _(max) = 1 - (283)/(373) = 0.24

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Now, on re designing collector, maximum temperature, T_(3max) changes to 300^(\circ), so, the new maximum efficiency,  \eta' _(max) is given by:

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The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles.


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