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Ostrovityanka [42]
3 years ago
7

Help with 1 2 and 3 please

Physics
1 answer:
geniusboy [140]3 years ago
6 0

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. I = \frac{q}{t} Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3.  So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

\frac{p1L1}{A1}  = \frac{p2L2}{A2}\\

We are looking for L2 so we can isolate using algebra to get:

\frac{A2(\frac{P1L1}{A1}) }{P2} = L2

If you fill in those values you get 0.0205

or 2.05 cm



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Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

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The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

      The  speed of the second  sphere is  v_1  =  20 \  ms^{-1}

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

      U  =  0.18 \  J

So

       Q =  0.3 +  0.18

       Q =  0.48 \  J

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

     KE_f  =  \frac{1 }{2}  m (v_2^2

substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

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