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LUCKY_DIMON [66]
2 years ago
11

3x/y=6g/b solve for x

Physics
1 answer:
lora16 [44]2 years ago
5 0
So looking at the problem, you are going to want to start by finding a common denominator (1) in this case: yb, and combining like terms (2). You are then going to want to multiply both sides by (yb) as the reciprocal to the fractions (3).
1)  3x    6g
     ---- = ---
     y       b

2)  3xb    6gy
     ------ = -----   
     yb       yb

3)       3xb    6gy
  (yb) ------ = -----  
          yb       yb
which becomes: 3xb = 6gy

So after this, things become much more simple, as all you have to do is isolate the (x), which can be done by dividing the entire equation by (3b).

3xb   6gy
----- = -----
3b      3b

where you will then find your answer of:
      2gy
x = -----       (simplified by the GCM of 3)
       b
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A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not
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A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is mov
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0.22m/s

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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current
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Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero  U = 0.1355 \ m

Explanation:

From the question we are told that

    The distance of wire one from two along the y-axis is    y = 0.340 m

   The current on the first wire is  I_1 =  (27.5i) A

    The force per unit length on each wire is  Z =  295 \mu N/m = 295*10^{-6}  N/m

Generally the force per unit length is mathematically represented as

         Z = \frac{F}{l}  =  \frac{\mu_o I_1I_2}{2\pi y}

=>      \frac{\mu_o I_1I_2}{2\pi y}  =  295

Where  \mu_o is the permeability of free space with a constant value of  \mu_o  =  4\pi *10^{-7} \ N/A2

substituting values

       \frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340}  =  295 *10^{-6}

=>    I_2 =  18.23 \ A

Let U  denote the  line in the xy-plane where the total magnetic field is zero

So  

      So the force per unit length of  wire 2  from  line  U is equal to the force per unit length of wire 1  from  line  (y - U)      

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substituting values

          \frac{  18.23  }{ U} =  \frac{ 27.5 }{(0.34 -  U) }

         6.198 -18.23U = 27.5U

          6.198=45.73U

          U = 0.1355 \ m              

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