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marshall27 [118]
2 years ago
15

An electric heater is madde of a wire of resistance 100π and connected to a 240v mains supply. Determine the power rating of the

heater​
Physics
1 answer:
nikklg [1K]2 years ago
5 0

Answer:

Power = 576 Watts

Explanation:

The electrical power of an electric circuit can be defined as a measure of the rate at which energy is either produced or absorbed in the circuit.

Mathematically, electrical power is given by the formula;

Electrical \; power = current * voltage

This ultimately implies that, the quantity (current times voltage ) is electrical power and it is measured (S.I units) in Watt (W).

Given the following data;

Resistance = 100 ohms

Voltage = 240 V

To find the power rating of the heater;

Power = V²/R

Where;

V is the voltage.

R is the resistance.

Substituting into the formula, we have;

Power = 240²/100

Power = 57600/100

Power = 576 Watts

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Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi
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Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}

Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

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