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Aliun [14]
3 years ago
15

1.5kj of energy are used every second by a microwave oven, what is the power rating of the oven?

Physics
2 answers:
Dennis_Churaev [7]3 years ago
7 0

Power is the rate at which energy is consumed. When 1 J of energy is consumed in one second, the power consumed is 1 watt.

The microwave consumes 1.5 kJ of energy.

In joules, the energy consumed is 1500 J

P=\frac{E}{t}

Here, E is the energy consumed in time interval <em>t.</em>

The power rating of the microwave is therefore,

P=\frac{1500J}{1 s} \\ =1500 W

Thus, the power rating of the oven is 1500 W or 1.5 kW

Artemon [7]3 years ago
7 0
Power = work/ time
Power = 1.5 kj/ 1s
Power = 1.5kW
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valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

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a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

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T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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Find more information: brainly.com/question/17203857

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