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OverLord2011 [107]
2 years ago
12

A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?

Physics
2 answers:
andreev551 [17]2 years ago
7 0

<span>Hello!
 
We have the following data:
</span>
Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

P =  \frac{E}{T}

Solving:

P = \frac{E}{T}

P = \frac{E}{T} \to T =  \frac{E}{P}

T =  \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}

\boxed{T = 3000\:seconds}

How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

\frac{1}{y} = \frac{60}{3000}

<span>Product of extremes equals product of means
</span>
60*y = 1*3000

60y = 3000

y =  \frac{3000}{60}

\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark


I hope this helps! =)
<span>

</span>
Vesnalui [34]2 years ago
6 0

Answer:

50 minutes

Explanation:

Power, P = 3 kW = 3000 W

Energy, E = 9 MJ = 9 x 10^6 J

Let the time be t.

Power = Energy / time

Time, t = E / P = 9 x 10^6 / 3000 = 3000 sec

t = 50 minutes

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Answer:

Explanation:

Given that,

A light bulb has a resistance of 2.9ohms

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And a battery of 1.5V is applied

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Rate of energy implies that we want to find power. Power is the rate at which work is done

P = Workdone / time

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In electronic, the power dissipated by a resistor is given as

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P = 1.5² / 2.9

P = 0.7759 W

P ≈ 0.776 W

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Explanation:

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Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

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\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

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Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

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Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

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