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nikklg [1K]
4 years ago
11

A wire has a current that flows to the right.

Physics
2 answers:
Reika [66]4 years ago
8 0

answer is DDDDDDDDDDDDDDDDDD


sergejj [24]4 years ago
8 0

D. circles around the wire that go in at the bottom

Explanation:

In order to determine the direction of the magnetic field around the wire, we should use the right's hand rule. First of all, let's keep in mind that the magnetic field around a current carrying wire consists of concentric circles around the wire, so the only possible options are C and D.

In the right-hand rule, we put our thumb in the same direction of the current (so, toward the right). Then we wrap all the other fingers of the hand around the thumb: the direction of the fingers give the direction of the magnetic field. In this case, the fingers go in at the bottom of the thumb (bottom of the wire), so the correct option is

D. circles around the wire that go in at the bottom

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Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

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To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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