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3 years ago

if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be

Physics

1 answer:

Kisachek [45]3 years ago

4 0

Answer:

$I_{rms}=2.12\ A$

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

$I=3\sin\theta\ A$

We need to find the rms value of the current.

The general equation of current is given by :

$I=I_o\sin\theta$

It means, $I_o=3\ A$

We know that,

$I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A$

So, the rms value of current is 2.12 A.

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PLEAASE HELP KE WITH THESE THREE YOULL GET POINTS

lozanna [386]

Answer:

0.0121
19.8387
2.2632

Explanation:

$\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{0.01223}{1.01}\:\\\mathrm{is}\:0.0121$

$\frac{\begin{matrix}\space\space&\textbf{\space\space}&\space\space&\space\space&\space\space&4&9&10\\ \space\space&\textbf{1}&9&.&8&\linethrough{5}&\linethrough{10}&\linethrough{0}\\ -&\textbf{0}&0&.&0&1&1&3\end{matrix}}{\begin{matrix}\space\space&\textbf{1}&9&.&8&3&8&7\end{matrix}}\\\\=19.8387$

$0.1886\times 12\\\\Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer\\1886\times\:12=22632\\\\0.1886\mathrm{\:has\:}4\mathrm{\:decimal\:places}\\12\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}4\mathrm{\:decimal\:places}\\\\=2.2632$

3 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

Irina18 [472]

Answer:

a. $q_C$ experiences the greatest net force and $q_B$ experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

$\displaystyle F=K\frac{q_1q_2}{d^2}$

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

$F_{TA}=-F_B-F_C+F_D$

Computing the individual forces we have

$\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}$

$\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}$

$\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}$

The total force on qA is:

$\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}$

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

$\displaystyle F_{TB}=F_A-F_C+F_D$

$\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}$

$\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

$\displaystyle F_{TC}=F_A+F_B+F_D$

$\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}$

$\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

$\displaystyle F_{TD}=-F_A-F_B-F_C$

$\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}$

$\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}$

Comparing the relative values of all the forces

$\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|$

This means that qc experiences the greatest net force and qB experiences the smallest net force

The ratio of the greatest to the smallest forces is

$\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9$

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3 years ago

Which energy changes take place when a pedaling cyclist uses a generator (dynamo) to light his bicycle

Fynjy0 [20]

When the pedaling cyclist uses a generator (dynamo) to light his bicycle lamp, the energy change is from mechanical to electrical.

<h3>Law of conservation of energy</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.

Thus, we can conclude the following as it relates to energy changes;

When the pedaling cyclist uses a generator (dynamo) to light his bicycle lamp, the energy change is from mechanical to electrical.

Learn more about conservation of energy here: brainly.com/question/166559

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