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kirill115 [55]
2 years ago
11

two long parrallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction.Find the magnitude of

theforce per length that each wires exerts on the other wire and indicates if the force is attractive or repulsive

Physics
1 answer:
Vera_Pavlovna [14]2 years ago
3 0

Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.

The force is repulsive.

Explanation: Please see the attachments below

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Of the following states of motion, the one which requires balanced forces is:
Tanzania [10]
I believe it would be D a change in direction of motion
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All of the outer planets are much larger than the inner planets true or false
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4 0
3 years ago
6) Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and blac
KIM [24]

Answer:

A) 1.5 v

B) Top plate is at higher voltage than the bottom plate

Explanation:

Battery value set between 0.0 V and 1.5 V

a) The potential difference between the plates

Δ V = V1( potential at top plate) - V2( potential at lower plate )

potential at top plate = 1.5 V

potential at lower plate = 0.0 V

hence potential difference = 1.5 V

b ) The top plate is always connected to the positive terminal of the DC source ( which is at a higher potential )while the bottom plate is connected to the negative terminal of the DC source ( which is at a lower potential )

hence the Top plate is at higher voltage than the bottom plate

7 0
3 years ago
Two teams are playing tug of war. Team A pulls to the right with a force of 450 N. Team B pulls to the left with a force of 415
motikmotik

Answer:

35 N to the right.

Explanation:

450 is going to the right so you subtract what is going against it. Which gives you 35. And because 450 is bigger than 415, it'll be going to the right.

6 0
2 years ago
A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads
Bond [772]

The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is calculated as follows;

Capacitors in series;

1/Ct = 1/8 + 1/7.5

1/Ct = 0.25833

Ct = 3.87 mF

Capacitors is parallel;

Ct = 3.87 mF + 12 mF + 15 mF

Ct = 30.87 mF

Ct = 0.03087 F

<h3>Charge in each capacitor</h3>

Q = CV

Q = 0.03087 x 12

Q = 0.37 C

Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

Learn more about capacitors here:  brainly.com/question/13578522

#SPJ1

3 0
1 year ago
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