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8090 [49]
3 years ago
9

If a resistor only has three color bands, it has a tolerance of 20 percent. What would this mean for a resistor specified at 10,

000 ohms?
A. It will fail, on average, on one out of five cycles, meaning its effective ohms is only 8,000.
B. It will fail if more than 12.000 volts pass through it
C. Its actual resistance, when measured, falls between 8k and 12 k ohms.
D. It allows 8000 volts to pass through it.
Engineering
1 answer:
Norma-Jean [14]3 years ago
3 0
B is the correct answe it will fail if more than 12.000
Hope this helps
:)
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A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}

  r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Now by putting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

  Ф=18.03°

So the shear angle is 18.03°.

4 0
4 years ago
a city with a large population is planning a bridge that will cross a river. The river is very deep and wide with a muddy bottom
Mrrafil [7]

Answer:

i think the city should not do it honestly it could cause way more traffic.

3 0
2 years ago
Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could
VladimirAG [237]

Answer:

4265.04\ \text{m}

2.38\times 10^{10}\ \text{W}

Explanation:

PE = Energy of food = 500 cal = 500\times4184=2.092\times10^6\ \text{J}

m = Mass of object = 50 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Potential energy of food is given by

PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}

Nancy could raise the weight to a maximum height of 4265.04\ \text{m}.

Mass of H_2 used per year = 25\times 10^{9}\ \text{kg/year}

Energy of H_2 = \dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}

Power

P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}

The power requirement is 2.38\times 10^{10}\ \text{W}.

6 0
3 years ago
A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl
lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

  • let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
  • let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
  • let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

  1. <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

          .<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

          .<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

            .<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

           .<em>5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = \\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )

NTU = \frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582   -1} )

NTU = 0.661

          <em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = \frac{NTU C{min} }{U}

As = \frac{0.661 x 5016 W. °c }{640 W/m²}

As = 5.18 m²

            <em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = \frac{As}{\pi D}

L = \frac{5.18 m² }{\pi (0.015 m)}

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0
3 years ago
a turbine operating at steady state at 500 kPa, 860 K and exists at 100 kPa. A temperature sensor indicates that the exit air te
noname [10]

Answer:

Given:

P₁ = 500 kPa

T₁ = 860 K

P₂= 100 kPa

T₂ = 460 K

Let's take entropy properties of T1 and T2 from ideal properties of air,

at T = 860K, s(T₁) = 2.79783 kJ/kg.K

at T = 460K, s(T₂) = 2.13407 kJ/kg.K

using entropy balance equation:

\frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}]

\frac{\sigma _cv}{m} = 2.79783 - 2.13407 - 0.287 In [\frac{100}{500}]

= - 0.2018 kJ/kg. K

In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).

8 0
3 years ago
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