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3 years ago

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the electromagnet shown in the image has a wire coiled 5 times around a nai. it can lift up to 7 metal paper clips. What two cha

nges would cause the electromagnet to increase its electromagnetic force and lift more paper clips?

2 answers:

stich3 [128]3 years ago

6 0

Answer: one of the answers should be increase the number of turns of wire on the nail

Explanation: the more you wrap the Wire on a nail, its electromagnet would increase

Dennis_Churaev [7]3 years ago

3 0

Answer:

Explanation:

The electromagnetic force of an electromagnet can be increased by:

1. wrapping the wire around the coil more times; or

2. increasing the current passing through the coil.

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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:

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

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$F = 25000 lb$

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:

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

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

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F = 25000lb + 150000lb

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$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

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3 years ago