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PIT_PIT [208]
2 years ago
14

Two charged particles, with charges q1 = q and q2 = 4q are located at a distance d = 2cm apart on the x axis A third charged par

ticle, with charge q3 = q is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 10n C
Physics
1 answer:
weqwewe [10]2 years ago
3 0

Answer:

0.667 cm

Explanation:

|F_1on3| = |F_2on3|

k×q1×q3/x^2 = k×q2×q3/(d-x)^2

q1/x^2 = q2/(d-x)^2

(d-x)^2/x^2 = q2/q1

(d-x)/x = sqrt(q2/q1)

(2 - x)/x = sqrt(4×q/q)

(2 - x)/x = 2

2 - x = 2x

2 = 3x

x = 2/3

x= 0.667 cm

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Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h
Neko [114]

Answer

given,

firm is producing  = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

\dfrac{MP_l}{MP_c}=\dfrac{40}{25}

\dfrac{MP_l}{MP_c}=\dfrac{8}{5}

\dfrac{W}{R}=\dfrac{20}{10}

\dfrac{W}{R}=2

\dfrac{MP_l}{MP_c} < \dfrac{W}{R}

Firm is not minimizing the cost because the firm use more capital and less labor.

3 0
3 years ago
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
Which of the following would increase​
bazaltina [42]

Answer:

1 and 3

Explanation:

<u>1 and 3  </u>

Increasing coils increases strength

   COOLING the wire would increase current flow and strength of magnet

Adding an iron core will definitely increase the strength of the electromagnet

7 0
2 years ago
Read 2 more answers
Which of the following is located in the stratosphere? (10 points) Weather, The ozone layer, UV rays, Birds.
sergij07 [2.7K]
The O Zone Layer
Hope I am right
5 0
3 years ago
Read 2 more answers
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep
Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

  m1 = 0.0225 Kg

  V1 = (0.869, -0.283) m / s

 

  Px = m Vx

  Px1 = 0.0225 0.869

  Px1 = 0.01955 Kg m

  Py = m Vy

  Py1 = 0.0225 (-0.283)

  Py1 = -0.006368 Kg m

  P1 = (0.0196, -0.00637) Kg m

Mouse 2

 m2 = 0.0223 Kg

 Px2 = 0.0223 (-0.883) = -0.0196 Kg m

 Py2 = 0.0223 (-0.253) = -0.00564 Kg m

 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

 m3 = 0.0197

 Px3 = 0.0197 0.345 = 0.00680 Kg m

 Py3 = 0.0197 0.803 = 0.0158 Kg m

 P3 = (0.00680, 0.0158) Kg m

Mouse4

  m4 = 0.0127 Kg

  Px4 = 0.0127 (-0.555) = -0.00705 Kg m

  Py4 = 0.0127 0.205 = 0.00260 Kg m

  P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

   Px = Px1 + Px2 + Px3 + Px4  

   Py = py1 + Py2 + Py3 + Py4

   Px = 0.0196 -0.0196 +0.00680 -0.00705

   Px = -0,000250 Kg m

   Py = -0.00637 -0.00564 +0.0158 +0.00260

   Py = 0.00639 Kg m

   P = (-0.000250, 0.00639) Kg m

7 0
3 years ago
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