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Artemon [7]
2 years ago
10

The heating element in a kettle behaves like a resistor. A particular kettle needs to operate at 230 V, with a power of 1500 W.

Calculate the resistance which the heating element needs to have​
Physics
1 answer:
dedylja [7]2 years ago
4 0

Answer:

R = 35.27 Ohms

Explanation:

Given the following data;

Voltage = 230V

Power = 1500W

To find the resistance, R;

Power = V²/R

Where:

V is the voltage measured in volts.

R is the resistance measured in ohms.

Substituting into the equation, we have;

1500 = 230²/R

Cross-multiplying, we have;

1500R = 52900

R = 52900/1500

R = 35.27 Ohms.

Therefore, the resistance which the heating element needs to have​ is 35.27 Ohms.

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pantera1 [17]

Answer:

A. The closest point in the Moon's orbit to Earth

Explanation:

The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.

Let's see instead the names of the other options:

B. The farthest point in the Moon's orbit to Earth  --> this point is called apogee

C. The closest point in Earth's orbit of the Sun  --> this point is called perihelion

D. The Sun's orbit that is closest to the Moon --> this point has no specific name

8 0
3 years ago
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anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

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initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

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This is a statement but yes a star forms inside nebulae which are gigantic clouds of gas. stars form inside as the gases own gravity pulls it together after which it becomes large enough to perform fusion and become a star.
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In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables
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