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Tanzania [10]
1 year ago
13

Two particles with oppositely signed charges nC are placed at two of the vertices of an equilateral triangle with side length 3

m. What is the magnitude of the electric field at the third vertex of the triangle
Physics
1 answer:
babymother [125]1 year ago
5 0

The magnitude of the electric field at the third vertex of the triangle is determined as zero.

<h3>Electric field at the third vertex of the triangle </h3>

The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;

E = E(13) + E(23)

E = (kq₁)/r² + (kq₂)/r²

where;

  • q1 is positive charge
  • q2 is negative charge

E =  (kq₁)/r² - (kq₂)/r²

E = 0

Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

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Nikolas had an idea that he could use the compressed carbon dioxide in a fire extinguisher to propel him on his skateboard.
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The Newton’s law Nikolas would use to come up with this idea is the <span>Third law that states:

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The average diameter of one tennis ball in a package of three is 6.8 cm. Which of the following is the combined volume of all th
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We want to find the combined volume of 3 tennis balls. We will get that the combined volume is 493.7 cm^3

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V = \frac{4}{3}*3.14*(\frac{D}{2})^3

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2 years ago
~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

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