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Lubov Fominskaja [6]
3 years ago
11

A 250 watt electric bulb is lighted for 5 hours daily and four 6 watt bulbs are lighted for 4.5 hours daily. Calculate the energ

y consumed (in kWh) in the month of February.
Physics
1 answer:
densk [106]3 years ago
5 0

Answer:

38.024 KWh

Explanation

250 watts multiplied with 5 hours daily multiplied with 28 days monthly equals 35000 Watt hours. Divided by a 1000 to get the KWh equals 35 KWh.

= 250×5hrs× 28days

= 35000watts

= 35000/1000

= 35KWh

Four 6 watt lightbulbs equals 24 watts 4x6=24

Hence, 24×4.5hrs×30days

= 3024watts

= 3024/1000

= 3.024KWhr

The total amount of energy consumed in the month of February = 35 KWh + 3.024 KWh = 38.024 KWh

Note that I had to use 28days since we are considering the month of February.

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A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

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10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

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3 years ago
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