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STatiana [176]
4 years ago
8

Two long, parallel wires are separated by a distance of 2.00 cm . The force per unit length that each wire exerts on the other i

s 3.60×10^−5 N/m , and the wires repel each other. The current in one wire is 0.640 A.
a. What is the current in the second wire?
b. Are the two currents in the same direction or in opposite directions?
Physics
1 answer:
RUDIKE [14]4 years ago
7 0

Answer:

a) current in the second wire is 5.60A

b) opposite directions

Explanation:

a) We need to find the current of wire, the magnitude of the force per unit length between the two wires carrying current I and I¹ is given by

\frac{F}{L} = \frac{U_0I^1}{2\pi r} \\\\I^1 = \frac{F2\pi r}{LU_0I}

= (3.6 * 10^-^5)\frac{2\pi (0.02)}{4\pi * 10^-^7)(0.64)} \\= 5.60A

b) knowing that for a two parallel conductor carrying current in the same direction attracts each other, and for a two parallel conductors carrying carying current in opposite direction repels eachother.

therefore, since the two wire repel each other then the current in the second wire must flow in the opposite direction of the current in the first wire.

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For this question we should apply
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A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma
scoundrel [369]

The gravitational force on the woman is A) 500 N

Explanation:

There are two forces acting on the woman during her fall:

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According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

\sum F=ma

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The net force can be written as

\sum F = F_G - F_D

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

a=0

Combining the equations together, we get:

F_G-F_D = 0

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

F_G=F_D=500 N

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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

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mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

6 0
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When you go for a walk which of the following forces is paired with the force of friction on your shoe
Ann [662]

Answer:

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