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Angelina_Jolie [31]
4 years ago
9

Identical point charges (+50 x 10 power -6C) are placed at the corners of a square with sides of 2.0-m length. How much external

energy is required to bring a fifth identical charge from infinity to the geometric center of the square?
Physics
1 answer:
Gnesinka [82]4 years ago
8 0

Answer:

636.4 J

Explanation:

The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C  and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)

Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r

U' = 4kq²/r

= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m

= 900 Nm²/√2 m

= 636.4 J

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Explanation:

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suppose the same amount of heat is applied to two bars. they have the same mass, but experience different changes in temperature
Andreyy89

If both bars are made of a good conductor, then their specific heat capacities must be different. If both are metals, specific heat capacities of different metals can vary by quite a bit, eg, both are in kJ/kgK, Potassium is 0.13, and Lithium is very high at 3.57 - both of these are quite good conductors.

If one of the bars is a good conductor and the other is a good insulator, then, after the surface application of heat, the temperatures at the surfaces are almost bound to be different. This is because the heat will be rapidly conducted into the body of the conducting bar, soon achieving a constant temperature throughout the bar. Whereas, with the insulator, the heat will tend to stay where it's put, heating the bar considerably over that area. As the heat slowly conducts into the bar, it will also start to cool from its surface, because it's so hot, and even if it has the same heat capacity as the other bar, which might be possible, it will eventually reach a lower, steady temperature throughout.

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4 years ago
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns
Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is  12 V

Answer:

The value  is  N_s  =  21 \  turns

Explanation:

From the equation we are told that

   The input voltage is  V_{in}  = 120 \ V

   The number of turns of the primary coil is N_p =  210 \  turn

    The output from the secondary is V_o =  12V

From the transformer equation

   \frac{N_p}{V_{in}}  =\frac{N_s}{V_o}

Here N_s is the number of turns in the secondary coil

=> N_s  =  \frac{N_p}{V_{in}}  *  V_s

=>N_s  =  \frac{210}{120}  *  12

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4 0
4 years ago
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3 years ago
A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of
spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

                                       Q =  m×c×ΔT

                        Heat gain by aluminium + heat lost by copper  = 0    (1)

For Aluminium:

      Q = m\times0.897\frac{J}{g.k}\times(25-5)

      Q = m x 17.94 joule

For Copper:

Q= 206g\times0.385\frac{J}{g.k} \times(88-25)

       Q= 4996.53 Joule

from eq 1

     m x 17.94 = 4996.53

     mass of aluminium = \frac{4996.53}{17.94} g

    Mass of the aluminium chunk = 278.51 g

                         

3 0
3 years ago
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