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victus00 [196]
3 years ago
11

Cell phones require powerful batteries in orde to work effectively. Which activity is best described as an engineering endeavor

related to cell phone batteries
Engineering
1 answer:
natta225 [31]3 years ago
8 0
Where are the options?
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Five kg of nitrogen gas (N2) in a rigid, insulated container fitted with a paddle wheel is initially at 300 K, 150 kPa. The N2 g
andrew-mc [135]

Answer:

A) attached below

B) 743 KJ

C) 1.8983 KJ/K

Explanation:

A) Diagram of system schematic and set up states

attached below

<u>B) Calculate the amount of work received from the paddle wheel </u>

assuming ideal gas situation

v1 = v2 ( for a constant volume process )

work generated by paddle wheel = system internal energy

dw = mCv dT .     where ; Cv = 0.743 KJ/kgk

     = 5 * 0.743 * ( 500 - 300 )

     = 3.715 * 200 = 743 KJ

<u>C) calculate the amount of entropy generated  ( KJ/K )</u>

S2 - S1 = 1.8983 KJ/K

attached below is the detailed solution

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3 years ago
A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL
prohojiy [21]

Answer:

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Explanation:

7 0
2 years ago
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Define the Static Balancing.
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Answer: Static Balancing is when a stationary object has the the ability to balance because its core centre of gravity is on the axis of rotation.

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A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de
Margaret [11]

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

  1. L = 20 cm d x 1 = 0.21 cm
  2. dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
  3. (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
  4. b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
  5. = 1561.84 MPa.

Read more about the diameter :

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2 years ago
Activity
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Answer:

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answer

Explanation:

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4 years ago
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