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hichkok12 [17]
3 years ago
15

_____ can be defined as the rate at which work is done or the amount of work done based on a period of time. (2 Points) voltage

power resistant current
Engineering
1 answer:
Ymorist [56]3 years ago
4 0

Answer: Power

Explanation:

The rate at which work is done or the amount of work done based on a period of time is referred to as power.

Power can also be defined as the amount of energy that is being transferred per unit time. The unit of power is one joule per second or simply called the watt.

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____________engineers build and test tools and machines, and they rely heavily on computer, math, and problem-solving skills.
Wittaler [7]

Answer:Mechanical Engineers

Explanation: Mechanical engineering is an engineering branch that combines engineering physics and mathematics principles with materials science to design, analyze, manufacture, and maintain mechanical systems. It is one of the oldest and broadest of the engineering branches.

8 0
3 years ago
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A strip of chicken skin was excised for mechanical testing in tension. The initial dimensions of the rectangular specimen were 3
Stells [14]

Answer:

Table and chart are attached below

Explanation:

4 0
4 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
AURORKA [14]

Answer:

a) w_{out} = 281.55\,\frac{kJ}{kg}, b) s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

-q_{out} - w_{out} + h_{in}-h_{out} = 0

w_{out} = h_{in} - h_{out}-q_{out}

w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}

w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}

w_{out} = 281.55\,\frac{kJ}{kg}

b) The entropy production is determined after the Second Law of Thermodynamics:

-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0

s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}

s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)

s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)

s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}

3 0
3 years ago
A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re
Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0
3 years ago
A discrete-time LTI system H has input x[n] and output y[n] related by the linear constant coefficient difference equation y[n]
Molodets [167]

Answer:

See explaination and attachment

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Have the solution as an attachment.

4 0
3 years ago
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