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Anettt [7]
3 years ago
8

Why is it important to cut all the way through an electrical wire on the first try?

Engineering
1 answer:
lbvjy [14]3 years ago
6 0

Answer:

I always thought it was so that the older wire could not have a problem and have another electrician must come back and fix it.

Explanation:

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Explain how collecting mean values (X-bar) from samples drawn from a dataset, regardless of the statistical distribution of the
Pani-rosa [81]

Answer:

answer is given below

Explanation:

Central Limit Theorem: The Central Limit Theorem (CLT) is a statistical theory that gives a sufficiently large sample size with limited variations from the population, the average of all samples from the same population is approximately the same. . In addition, all models follow a nearly normal distribution model.

The given phenomenon is described in the central limit theory. In other words, if we repeatedly take independent random samples of size n from any population, when n is large, the sample distribution is the normal distribution pattern.

mean of the sample means

                    \mu _\bar x = \mu        .............1

and here standard deviation of the sample means is

                      \mu _\bar x = \frac{\sigma }{\sqrt{n}}      .........2

and This theory is found elsewhere in the field of statistics. Although the central limit theory may seem abstract and devoid of any application, this theory is actually important for statistical practice.

8 0
4 years ago
Write what you already know about college majors. What are they? Can you think of any examples? When do you have to pick one? Ca
Mandarinka [93]
College majors are specific fields of study that help people prepare for career paths and learn content related to that subject. Some college majors would criminal justices, forensic science, gender studies, engineering, chemistry, and more. You should generally begin to research in your junior year of high school and select one by the beginning or your senior year for college applications. Once you get to college you can change your major.
5 0
3 years ago
Read 2 more answers
Identify the unit of the electrical parameters represented by L and C and prove that the resonant frequency (fr)=1÷2π√LC
Gre4nikov [31]

Answer:

L = Henry

C = Farad

Explanation:

The electrical parameter represented as L is the inductance whose unit is Henry(H).

The electrical parameter represented as C is the inductance whose unit is Farad

Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :

To obtain :

At resonance, Inductive reactance = capacitive reactance

Equate the inductive and capacitive reactance

Inductive reactance(Xl) = 2πFL

Capacitive Reactance(Xc) = 1/2πFC

Inductive reactance(Xl) = Capacitive Reactance(Xc)

2πFL = 1/2πFC

Multiplying both sides by F

F * 2πFL = F * 1/2πFC

2πF²L = 1/2πC

Isolating F²

F² = 1/2πC2πL

F² = 1/4π²LC

Take the square root of both sides to make F the subject

F = √1 / √4π²LC

F = 1 /2π√LC

Hence, the proof.

8 0
3 years ago
How are radio waves carried?
bagirrra123 [75]

Answer:

c from transmitter to a receiver

6 0
4 years ago
Read 2 more answers
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
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