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nekit [7.7K]
3 years ago
11

2O POINTS!!! Easy Question. Will Mark Brainiest

Physics
1 answer:
wel3 years ago
6 0

Answer: The Electrostatic force of attraction or repulsion between two charges shows that the Newton's third law applies to electrostatic forces.

Explanation: Consider two Oppositely charged charges separated by distance d.

The electrostatic force exerted by charge 1 on charge 2 is.

By Coulomb's Law :

               F1  = k \frac{Q1 Q2}{d2}.....................................(1)

The electrostatic force exerted by charge 2 on charge 1 is.

              F2  = - k \frac{Q1 Q2}{d2} ................................. (2)

negative sign shows that force are in opposite direction.

 

From Equation 1 and 2

              F1 = - F2

Which implies Newton Third law.

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Answer:

a) E = 6.024\,USD, For m kilograms, it is 4184m J., 3600000 joules, b) i = 88.200\,A

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})

Where:

m_{w} - Mass of water, measured in kilograms.

c_{w} - Specific heat of water, measured in \frac{J}{kg\cdot ^{\circ}C}.

T_{f}, T_{i} - Initial and final temperatures, measured in ^{\circ}C.

Then,

Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)

Q_{needed} = 180748800\,J

The energy needed in kilowatt-hours is:

Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)

Q_{needed} = 50.208\,kWh

The electric energy required to heat up the water is:

E = \frac{50.208\,kWh}{0.75}

E = 66.944\,kWh

Lastly, the cost of heating a hot tub is: (USD - US dollars)

E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)

E = 6.024\,USD

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}

i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}

i = 88.200\,A

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