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fredd [130]
3 years ago
7

A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field h

as a value of 40.0 N/C directed to the south. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?
Physics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

E_total = - 50 N / A

Explanation:

The electric field is a vector magnitude whereby

          E_total = Eₐ + E_b

where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line

         E_total = - E_a - E_b

         

the electric field for a point charge is

        E_a = k \ \frac{q_a}{r_a^2 }

        qₐ= Eₐ rₐ² / k

indicates that Eₐ = 40.0 N / C

        qₐ = 40.0 0.250²/9 10⁹

        qₐ = 2.777 10⁻¹⁰ C

indicates that the charge of the two points is the same

        qₐ = q_b

      E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²

      E_total = -k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )

       

we calculate

       E_total = - 40.0 (5/4)

       E_total = - 50 N / A

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Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

                    = 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0
3 years ago
If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a
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Answer:

500J

Explanation:

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Two long, straight wires are parallel and 26 cm apart.
mezya [45]

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

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3 years ago
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