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Helga [31]
4 years ago
15

Two long, parallel wires are attracted to each other by a force per unit length of 305 µN/m. One wire carries a current of 25.0

A to the right and is located along the line y = 0.470 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
Physics
1 answer:
kykrilka [37]4 years ago
5 0

To solve this problem we will use the concepts related to the electromagnetic force related to the bases founded by Coulumb, the mathematical expression is the following as a function of force per unit area:

\frac{F}{L} = \frac{kl_1l_2}{d}

Here,

F = Force

L = Length

k = Coulomb constant

I =Each current

d = Distance

Force of the wire one which is located along the line y to 0.47m is 305*10^{-6}N/m then we have

l_2 = \frac{F}{L} (\frac{d}{kl_1})

l_2 = (305*10^{-6}N/m)(\frac{0.470m}{(2*10^{-7})(25A))})

l_2 = 28.67A

Considering the B is zero at

y = y_1

\frac{kI_2}{2\pi y} =\frac{kI_1}{2\pi y_1}

\frac{(4\pi*10^{-7})(28.67)}{2\pi (y_1)} = \frac{(4\pi *10^{-7})(25)}{2\pi (0.47-y_1)}

y_1 = 0.25m

Therefore the value of y for the line in the plane of the two wires along which the total B is zero is 0.25m

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kakasveta [241]

Answer:

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Explanation:

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4 0
4 years ago
A bullet is fired horizontally into an initially stationary block of wood suspended by a string and remains embedded in the bloc
UkoKoshka [18]

The ratio of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet is 0.78 %.

<h3>Final velocity of the block/bullet system</h3>

Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

where;

  • h is the maximum height reached by the system
  • v is the initial velocity of the system

v = √(2 x 9.8 x 1.1)

v = 4.64 m/s

<h3>Initial velocity of the bullet</h3>

Apply the principle of conservation of linear momentum.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

  • u₁ is the initial velocity of the bullet
  • u₂ is the initial velocity of the block
  • v is the final velocity after collision
  • m₁ is mass bullet
  • m₂ is mass of block

(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)

0.0075u₁ = 4.4428

u₁ = 4.4428/0.0075

u₁ = 592.37 m/s

<h3>Initial kinetic energy of the bullet</h3>

K.Ei = ¹/₂m₁u₁²

K.Ei = ¹/₂(0.0075)(592.37)²

K.Ei = 1,315.88 J

<h3>Final kinetic energy of the block/bullet system</h3>

K.Ef = ¹/₂(m₁ + m₂)v²

K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²

K.Ef = 10.31 J

<h3>Ratio of final kinetic energy to initial kinetic energy</h3>

= K.Ef/K.Ei x 100%

= (10.31 / 1,315.88) x 100%

= 0.78 %

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

4 0
2 years ago
A wave's _______________ and its frequency have a reciprocal relationship.
Varvara68 [4.7K]
Wavelength and frequency have a reciprocal relationship. If one doubles, the other halves.
8 0
4 years ago
Read 2 more answers
Suppose high tide is at midnight, the water level at midnight is 3 m, and the water level at low tide is 0.5 m. Assuming the nex
aev [14]

We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

t=10.0hours

From the Question we are told that

Maximum height h_{max}=3m

Minimum height  H_{min}=0.5m

Time for  next high tide will occurT=12 hours =>720 min

Generally Average Height

h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75

Therefore determine Amplitude to be

A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25

Generally, the equation for Time is mathematically given by

At t=0

h(x)=Acos(Bx)+h_{avg}

Where

B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}

Therefore

h(t)=Acos8.73*10^{-3}(t)+h_{avg}

Hence the Time at T=1.125 is

1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75

-0.1249t=1.75

t=10.0hours

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5 0
3 years ago
Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a
I am Lyosha [343]

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

q_1\ and\ q_2 are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

F'=\dfrac{kq^2}{r'^2}

F'=\dfrac{kq^2}{(2r)^2}

F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}

F'=\dfrac{1}{4}\times 7.35

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

3 0
3 years ago
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