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Zinaida [17]
3 years ago
15

You are cooking breakfast for yourself and a friend using a 1,140-W waffle iron and a 510-W coffeepot. Usually, you operate thes

e appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period
Physics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

The cost is 297 cents.

Explanation:

Power of iron, P = 1140 W

Power of coffee pot, P' = 510 W

Voltage, V = 110 V

Time, t = 0.5 h each day

Cost = 12 cents per kWh

(a) Total energy

E = P x t + P' x t

E = 1140 x 0.5 x 60 x 60 + 510 x 0.5 x 60 x 60

E = 2052000 + 918000 = 2970000 J

1 kWh = 3.6 x 10^6 J

E = 0.825 kWh

For 30 days

E' = 0.825 x 30 = 24.75 kWh

So, the cost is

= 12 x 24.75 = 297 cents

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Answer:

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Explanation:

If

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we can apply the equation:

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then we use the equation of Time of Flight:

t = 2*V₀y / g   ⇒    V₀y = g*t / 2 = (9.81 m/s²)*(5 s) / 2  ⇒    V₀y = 24.525 m/s

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