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3 years ago

How much energy is transferred electrically by a 2000 W cooker in half an hour?

Physics

2 answers:

Kryger [21]3 years ago

5 0

Responder:

E = 1440 kJ

Explicación:

Se da que,

La potencia de un horno de cocción es de 800 W

El voltaje al que se opera es de 230 V

Tiempo, t = 30 minutos = 1800 segundos

Necesitamos encontrar la energía eléctrica utilizada por el horno de cocción. El producto de la potencia y el tiempo es igual a la energía consumida. Entonces,

katovenus [111]3 years ago

5 0

It's not transferred, it's consumed (converted into heat).
If the cooker is full on (all the plates at full), then 1kWh. Else it depends on the duty cycle. If a device is rated 2000W, that's the most it can consume, not what it will actually consume in normal use.

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ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

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Therefore on solving the above differential equation we get,

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where A(t)=A₀ $e^{\frac{-bt}{2m}}$

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$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

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Luda [366]

Answer:

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Explanation:

The first thing will be to draw a figure showing the condition,

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Regards M.Y.

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