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)

Answers

Answer 1
Answer:

Answer:

0.556 Watts

Explanation:

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

m=(w)/(g)=(762)/(9.81)

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)

Power

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.

Answers

Answer:

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)

Given:

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

Solution:

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

Now,

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)

Now,

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

where

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
• Draw a sketch of the reinforced concrete beam showing all dimensions, number and size of rebars, including stirrups.

Answers

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

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

Thus;

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"

Thus;

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

Hence;

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?

Answers

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

Chain processes

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

Throughout manufacturing, a six sigma process has been utilized just to generate a product throughout which 99.99966 percent among all possibilities to produce certain aspects of a part seem to be likely toward being defect-free.

Thus the response above is correct.

Find out more information about chain processes here:

brainly.com/question/25646504

Answer: The furthest upstream process that has problems.

A process in manufacturing is a particular method used for producing a product.

A six sigma process is used in processing to produce a product that is 99.99966% of all opportunities to produce some feature of a part are statistically expected to be free of defects.

According to the rules of the six sigma process, when there's a defect, the best thing to do is investigate the furthest upstream process that has problems.

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?

Answers

Answer:

\eta _(max) = 0.2413 = 24.13%

\eta' _(max) = 0.5061 = 50.61%

Given:

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

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

T_(3max) = 300^(\circ) = 273 + 300 = 573 K

Solution:

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

\eta _(max) = 1 - (T_(2))/(T_(1max))

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

\eta _(max) = 0.2413 = 24.13%

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

\eta' _(max) = 1 - (T_(2))/(T_(3max))

\eta _(max) = 1 - (283)/(573) = 0.5061

\eta _(max) = 0.5061 = 50.61%

What is the difference between absolute and gage pressure?

Answers

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

P_(ab)=P_(g)+P_(atm)

Here, P_(ab) is absolute pressure, P_(g) is gauge pressure andP_(atm) is atmospheric pressure.

Step2

Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.  

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.

Answers

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: False.

Safety risks can be defined as an assessment of the risks and occupational hazards associated with the use, operation or maintenance of an equipment or automobile vehicle that is capable of leading to the;

  • Harm of a worker (technician).
  • Injury of a worker (technician).
  • Illness of a worker (technician).
  • Death of a worker (technician).

Hybrid electric vehicles (HEVs) or EVs are typically designed and developed with parts or components that operates through the use of high voltageelectrical systems ranging from 100 Volts to 600 Volts. Also, these type of vehicles have an in-built HEV batteries which are typically encased in sealed shells so as to mitigate potential hazards to a technician.

On the other hand, conventional gasoline vehicles are typically designed and developed with parts or components that operates on hydrocarbon such as fuel and motor engine oil. Also, conventional gasoline vehicles do not require the use of high voltage electrical systems and as such poses less threat to technicians, which is in contrast with hybrid electric vehicles (HEVs) or EVs.

This ultimately implies that, the safety risks for technicians who work on hybrid electric vehicles (HEVs) or EVs are different from those who work on conventional gasoline vehicles due to high voltage electrical systems that are being used in the former.

In conclusion, technicians who work on hybrid electric vehicles (HEVs) or EVs are susceptible (vulnerable) to being electrocuted to death when safety risks are not properly adhered to unlike technicians working on conventional gasoline vehicles.

Find more information: brainly.com/question/2878752

Answer:

Batteries are safe when handled properly.

Explanation:

Just like the battery in your phone, the battery in some variant of an electric car is just as safe. If you puncture/smash just about any common kind of charged battery, it will combust. As long as you don't plan on doing anything extreme with the battery (or messing with high voltage) you should be fine.

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