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andrew-mc [135]

3 years ago

8

guys be careful noblepenguin is getting people banned he took all my brainliest he has to be stopped has a bot a be careful Kati

e as well

2 answers:

Julli [10]3 years ago

7 0

Answer:

K thanks for letting everyone know!

Explanation:

Veseljchak [2.6K]3 years ago

3 0

Thanks for letting us know

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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.

a.

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

b.

The ratio of the greatest to the smallest forces is

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

5 0

3 years ago

FILL IN TUE BLANK WITH ANSWER DOWN BELOW!!! When storm clouds produce lightning and thunder, 1)_________ energy changes to2)____

kkurt [141]

Electric potential, Chemical , sound

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3 years ago

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The sound level in a certain seat at the movie theater is higher than the level in the surrounding seats. Which wave interaction

Grace [21]

Answer: constructive interference

Explanation: plz mark Brainly

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Zorn and Porsha are ice skating. Porsha has a mass of 60 kg, and Zorn has a mass of 40 kg. As they face each

algol13

Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Missing part of the question: determine the magnitude of Porsha's acceleration.

Given the data in the question;

Mass of Porsha; $m_{porsha} = 60kg$

Mass of Zorn; $m_{zorn} = 40kg$

Force of Porsha push; $F_{porsha} = 168N$

Magnitude of Porsha's acceleration; $a = \ ?$

To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:

$F = m*a$

Where m is the mass of the object and a is the acceleration.

We substitute the mass of Porsha and the force he used into the equation

$168N = 60kg * a\\\\a = \frac{168kg.m/s^2}{60kg}\\\\a = 2.8m/s^2$

Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Learn more: brainly.com/question/25125444

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3 years ago

How does space promote science education

Fittoniya [83]

<span>NASA and the Mad Science Group of Montreal, Canada, have teamed in an effort to spark the imagination of children, encouraging more youth to pursue careers in science, technology, engineering and math. The two organizations recently signed a Space Act Agreement, officially launching the development of the Academy of Future Space Explorers.</span>

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