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2 years ago

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Q1. In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the

flow of electrons. What is the average number of electrons per second that flow past a fixed reference cross section that is perpendicular to the direction of flow? (5 Points)

1 answer:

Anit [1.1K]2 years ago

7 0

Answer:

2.9*10^14 electrons

Explanation:

An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.

An electron has a charge of 1.6*10^-19 C.

35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons

So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.

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You want to determine whether the race of the defendant has an impact on jury verdicts. You assign participants to watch a trial

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Answer:

The confidence scale represents an ordinal scale of measurement

Explanation:

An ordinal scale or level of measurement is used to measure attributes that can be ranked or ordered, but the interval between the attributes do not have quantitative significance. In this case, the measurement was done on a scale of 1 - 7, with a "1" being; not all that race of defendant has an impact on jury verdicts and a "7" being "very" meaning that race indeed has impact on jury verdicts. Another example can be a survey carried out on the level of customer satisfaction on a particular product, with "1" most dissatisfied and "10 " representing most satisfied. In the first example, it is wrong to say that the difference between 1 being "not at all" and maybe 3 is the same as the difference between 5 and 7 which have different connotations, because the numbers are merely for tagging and not to quantify.

Other levels of measurement include:

1. Nominal: this is the simplest level of measurement and it is simply used to categorize the attributes. Example is taking a survey on gender in the categories of male, female and transgender.

2. Interval: the interval scale is used when the distance between two attributes have meanings but there is no true zero value associated with the scale.

3. Ratio: this combines all the other three levels of measurement and is used to categorize, used to show ranking, has meaningful distances between the attributes and the scale has a true zero point. Example is the measurement of temperature using the celcius scale thermometer, where there is a true zero point at 0°C and the distance between 5°C and 10°C is the same as the distance between 10°C and 15°C.

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2 years ago

Two kilograms of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m3 and 1 bar, re

valentinak56 [21]

Answer: Heat transfer (Q) is 521 kJ.

Explanation: In the piston-cilinder assembly, we can suppose that the oxygen act as an ideal gas, so, it can be used the General Gas Equation:

PV= $\frac{m}{M}$ RT, where:

P is pressure;

V is volume;

m is mass;

M is molar mass;

R is a constant: R = 8.314. $10^{-5}$ m³bar.K⁻¹.mol⁻¹;

T is temperature;

Using this equation, find the intial temperature:

PV = $\frac{m}{M}$ RT

1.2 = $\frac{2}{16}$ .8.314. $10^{-5}$ .T

T = 1.924. $10^{5}$ K

To determine the final temperature, use Combined Gas Law:

$\frac{P_{i} . V_{i} }{T_{i} }$ = $\frac{P.V }{T}$ , in which, the left side of the equality is related to the initial values and the right side, to the final values.

As pressure is constant:

$\frac{V_{i} }{T_{i} } =\frac{V}{T}$

T = $\frac{V.T_{i} }{V_{i} }$

T = $\frac{4.1.924.10^{-5} }{2}$

T = 3.85. $10^{5}$ K

With the temperatures, calculate the heat transfer of the process:

Q = m.k.ΔT, where:

k is heat constant

ΔT = T - $T_{i}$

Q = m.k.ΔT

Q = 2.1.35.(3.85 - 1.92). $10^{5}$

Q = 521 kJ

The heat transfer in the process is 521 kJ.

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2 years ago

Read 2 more answers

Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t

alukav5142 [94]

Answer:

hoop stress
longitudinal stress
material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = $\frac{PD}{2T}$ EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = $\frac{PD}{4T}$ EQUATION 2

where p = water pressure inside the hose

d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

$\frac{D}{T}$ ≥ 20

insert this value into equation 1

αh = $\frac{20 *30}{2}$ = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

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2 years ago

A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz

nikitadnepr [17]

Answer:

25 - $\sqrt[4]{26.66*10^{-8} }$ mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2 $\pi$ f = 2 * $\pi$ * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter using the relation

$\frac{J}{c_{2} } = \frac{T}{t_{max} }$ = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

$\frac{J}{c_{2} }$ = $\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )$ = $\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ]$

therefore; 0.025^4 - $c^{4} _{1}$ = 0.050 / $\pi$ (7.8 *10^-6)

$c^{4} _{1}$ = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

$c_{1} = \sqrt[4]{26.66 * 10^{-8} }$ =

THE TUBE THICKNESS

$c_{2} - c_{1}$ = 25 - $\sqrt[4]{26.66*10^{-8} }$ mm

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2 years ago

Oliver is designing a new children’s slide to increase the speed at which a child can descend. His first design involved steel b

AVprozaik [17]

Answer:

The correct option is;

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Explanation:

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Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact

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