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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 pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor
Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm  

Hence, 1.16 cm were cut from the end of the second pipe

6 0
3 years ago
Pls help! What would be considered an antonym for law of superposition??
OverLord2011 [107]

Answer:

do you have any pictures or any answers we can choose from?

Explanation:

This is an incomplete question.

8 0
3 years ago
A small block of mass M = 0.10 kg is released from rest at point 1 at a height H = 1.8 m above the bottom of a track, as shown i
Ede4ka [16]

Answer:

D

Explanation:

A) is not correct, because the gravitation potential energy will depend on the height the block is located at. It will be calculated with the formula:

U=mgh.

If we take the ground as a zero height reference, then on point 2 the potential energy will be:

U_{2} = 0.10kg(9.81 m/s^{2})(0.6m)

U_{2}=0.59 J

While on point 3, the potential energy will be greater.

U_{3}=0.10kg(9.81 m/s^{2})(1.2m)

U_{3}=1.18 J

B) is not the right answer because the kinetic energy will vary with the height the block is located at in the fact that the energy is conserved (this is if we don't take friction into account or air resistance) so in this case:

U_{2}+K_{2}= U_{3}+K_{3}

We already know the potential energy at point 2. We can calculate the kinetic energy at point 3 like this:

K_{3} =\frac{1}{2}mv_{3}^{2}

K_{3} =\frac{1}{2}(0.10kg)(2.5 m/s)^{2}

K_{3} =0.31 J

So the kinetic energy at point 2 is given by the equation:

K_{2}  =U_{3}-U_{2}+K_{3}

so:

K_{2} = (1.18J)-(0.59J)+0.31J

K_{2} =0.9J

As you may see the kinetic energy at point 2 is greater than the kinetic energy at point 3.

C) Is not correct because according to the first law of thermodinamics, energy is not lost, only transformed. So, since we are not taking into account friction or any other kind of loss, then we can say that the amount of mechanical energy at point 1 is exactly the same as the mechanical energy at point 3.

D) Because of what we talked about on part C, this will be the true situation, because the mechanical energy of the block will be the same no matter theh point you measure it at.

7 0
3 years ago
6. Your 60-watt light bulb is plugged into a 110-volt household outlet and left on for 3 hours. The utility company charges you
elena55 [62]

Answer:

0.0198 dollars

Explanation:

We first obtain the total energy drawn by the light bulb in kilowatt-hour (kW-hr) using the following relationship;

Energy = power rating x time

Given;

power rating = 60W,

time = 3hrs

Therefore,

Energy consumed = 60W x 3hr

Energy consumed = 180W-hr = 0.18kW-hr

We then multiply this energy in kW-hr by he amount charged per unit kW-hr as follows;

Given that 1kW-hr= 0.11 dollars

0.18kW-hr = 0.18 x 0.11

                = 0.0198 dollars

6 0
3 years ago
Read 2 more answers
Which observation supports a model of the nature of light in which light acts as a wave?
Lorico [155]

Answer: A

Explanation:

5 0
3 years ago
Read 2 more answers
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