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3 years ago

Suppose the sphere is positively charged. Is it attracted to, repelled by, or not affected by the magnet?.

1 answer:

Nutka1998 [239]3 years ago

4 0

When a positively charged sphere is brought near the north pole of a magnet, the positively charged sphere will be attracted to the magnet.

<h3>Positively charged object</h3>

When a positively charged object is brought near the north pole of a magnet, the positively charged object will be attracted to the magnet beacuse of polarity.

Positively charged metals have the tendency to show the polarization of charges.

Thus, when a positively charged sphere is brought near the north pole of a magnet, the positively charged sphere will be attracted to the magnet. Also, if the south pole is brought near the sphere, the positively charged sphere will repel the magnet.

Learn more about attraction of magnet here: brainly.com/question/14749231

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A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpt

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A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

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Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$