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3 years ago

7

When are magnets attracted and when are they repelled?

2 answers:

ohaa [14]3 years ago

6 0

<em>when </em><em>South </em><em>and </em><em>North </em><em>poles </em><em>are </em><em>placed </em><em>together </em><em>they </em><em>attract </em><em>as </em><em>they </em><em>are </em><em>opposite</em><em> </em><em>poles. </em><em>In</em><em> </em><em>magnets </em><em>opposite </em><em>po</em><em>l</em><em>es </em><em>are </em><em>attracted </em><em>and </em><em>like </em><em>poles </em><em>are </em><em>repelled. </em>

yarga [219]3 years ago

6 0

Magnets are attracted when easy of the different sides, most commonly known as “north” and “south”, are facing eachother. They repel when north and north, or south and south are facing eachother.

hope this helps! :)

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Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

$W_s = \frac{1}{2}kc^2$

Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

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3 years ago

9. An object starting with a velocity of 20m/s accelerates to

Aliun [14]

Answer:

6.25 s

Explanation:

V=u+at

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3 years ago

Field lines point straight out from all sides of an object. Which statement best describes the object?

Mashcka [7]

It is positively charged.

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3 years ago

Read 2 more answers

A charge of 80 coulombs passes through a circuit in 5 seconds. What is the current through the circuit?

UkoKoshka [18]

Answer:

I = 16amp

Explanation:

Charge coulomb ( Q ) = It

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3 years ago

A positive charge is fixed at point (−4,−3) and a negative charge − is fixed at point (−4,0). Determine the net electric force ⃗

Masteriza [31]

The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

F = kq₁q₂/r²

where;

k is coulomb's constant
q₁ and q₂ are the charges
r is the distance between the charges

<h3>Distance between the charges</h3>

$|r| = \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2} \\\\|r| = \sqrt{(-4--4)^2+ (0--3)^2} \\\\|r| = \sqrt{(0)^2 + (3)^2} \\\\|r| = 3 \ units$

$F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \ N$

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

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2 years ago