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4 years ago

6

In the combination of resistors above, consider the 1.50 µΩ and 0.75 µΩ. How can you classify the connection between these two r

esistors?

1 answer:

Airida [17]4 years ago

3 0

Answer: they are connected in series.

Explanation:

You might be interested in

1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se

Lorico [155]

Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]² (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D² (3)

Where:

<em>t</em> - Cooling time of the cylindrical casting, in time unit.
<em>t'</em> - Cooling time of the prism casting, in time unit.
<em>C</em> - Cooling factor, in time unit per square meter.
<em>D</em> - Diameter of the cylinder, in meters.
<em>L</em> - Length of the casting, in meters.
<em>T</em> - Width of the cross section of the prism casting, in meters.

If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:

$T = \sqrt{\frac{\pi}{12} }\cdot D$

<em>T ≈ 0.153 m</em>

<em />

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C (2b)

Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:

<em>r = (</em>1 <em>- t'/t) ×</em> 100 %

<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %

<em>r =</em> 93.374 %

The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting. $\blacksquare$

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

3 0

3 years ago

Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent

monitta

Answer:

$F_r = 200N$

Explanation:

Given

Let the two forces be

$F_1 = 130N$

$F_2 = 110N$

and

$\tan(\theta) = \frac{12}{5}$

Required

Determine the resultant force

Resultant force (Fr) is calculated using:

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

This means that we need to first calculate $\cos(\theta)$

Given that:

$\tan(\theta) = \frac{12}{5}$

In trigonometry:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

By comparing the above formula to $\tan(\theta) = \frac{12}{5}$

$Opposite = 12$

$Adjacent = 5$

The hypotenuse is calculated as thus:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 12^2 + 5^2$

$Hypotenuse^2 = 144 + 25$

$Hypotenuse^2 = 169$

$Hypotenuse = \sqrt{169$

$Hypotenuse = 13$

$\cos(\theta)$ is then calculated using:

$\cos(\theta)= \frac{Adjacent}{Hypotenuse}$

$\cos(\theta)= \frac{5}{13}$

Substitute values for $F_1$ , $F_2$ and $cos(\theta)$ in

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

$F_r^2 = 130^2 + 110^2 + 2*130*110*\frac{5}{13}$

$F_r^2 = 16900 + 12100 + 11000$

$F_r^2 = 40000$

Take square roots of both sides

$F_r = \sqrt{40000$

$F_r = 200N$

<em>Hence, the resultant force is 200N</em>

4 0

3 years ago

A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and

ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

$h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg$

The quality at state 4 is determined from the condition $s_{4} =s_{3}$ and the entropies of the components at the condenser pressure taken from table:

$q_{4} =\frac{s_{4}-s_{liq50} }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935$

The enthalpy at state 4 then is:

$h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\$

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:

$h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg$

The enthalpy at state 2 is determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349$

Part B

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:

$h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg$

The enthalpy at state 2 is then determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=299.79 kJ/kg

The thermal efficiency in this case then is:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345$

8 0

3 years ago

Identify factors that can cause a process to become out of control. Give several examples of such factors.

Oliga [24]

Answer:

Explained

Explanation:

This situation can occur because of various factors such as:

Gradual deterioration of lubrication and coolant.
change of environmental condition such as temperature, humidity, moisture, etc.
Change in the properties of incoming raw material
An increase or decrease in the temperature of the heat treating operation
Debris interfering with the manufacturing process.

4 0

3 years ago

Which claim does president Kennedy make in speech university rice ?

mafiozo [28]

Answer: The United States must lead the space race to prevent future wars.

Explanation: Hope this helps

4 0

3 years ago

Read 2 more answers