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1 year ago

How does the government involve in a child or young person’s upbringing in your country?

1 answer:

8 0

The government are involved in a child or young person’s upbringing in my country by providing the right amenities and ensure the justice system is fair.

<h3>What is Upbringing?</h3>

This is defined as the way an individual treated at a young age by older adults such as parents.

Providing amenities will ensure the upbringing of the child is easier and a fair justice system will help to control crime.

Read more about Upbringing here brainly.com/question/12397063

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1. Which word or phrase best describes entropy?

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Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

5 0

3 years ago