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nikklg [1K]
3 years ago
11

A wire has a current that flows to the right.

Physics
2 answers:
Reika [66]3 years ago
8 0

answer is DDDDDDDDDDDDDDDDDD


sergejj [24]3 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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-- The string is 1 m long.  That's the radius of the circle that the mass is
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-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .

-- Centripetal acceleration is  V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²

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3 0
3 years ago
What is a half-life?
Serggg [28]

Answer:

option D is correct

Explanation:

5 0
2 years ago
Read 2 more answers
Consider a circuit with a main wire that branches into two other wires. If the current is 10 A in the main wire and 4 A in one o
k0ka [10]

If the primary wire's power is 10 A and one branch's power is 4 A, another branch's power will be 6A.

According to Kirchhoff's current law (KCL), the total current flowing through a parallel route circuit's junction equals the total current flowing away from it.

Provided that one of the two branches through which power exits the intersection has a flow of 4A, and also that the junction's overall flow entering it is 10A, the entire current going the junction should be 10A.

Consequently, the second wire's power may be expressed as;

I = I1+ I2 [ where I= total current (10A);

                I1= current in one branch (4A) &

                I2= current in another branch]

⇒I2 = I - I1

⇒I2 = 10A - 4A

⇒I2 = 6A

Therefore, it can be concluded that when the primary wire bears 10A power having 4A in one of its branches, another branch carries 6A power.

Learn more about Kirchhoff's law here:

brainly.com/question/6417513

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4 0
1 year ago
I need help guys it’s very confusing
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2 years ago
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