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Svetradugi [14.3K]
3 years ago
8

In a house, 3 bulbs of 60 watt each are lighted for 3 hours daily, 4 fans of 100 watt each are used for 8 hours daily and an ele

ctric heater of 2 kwh used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 4/kwh.​
Physics
1 answer:
Virty [35]3 years ago
5 0

Answer:

Total Energy consumed = 146.94 kwh

Total cost = rs 587.76

Explanation:

Given:

3 bulbs each 60 watt [for 3 hours daily]

4 fans each 100 watt [for 8 hours daily

1 electric heater of 2 kWh (2,000 watt) [for half hour daily ]

Rate = rs 4/Kwh

Find:

Total Energy consumed

Total cost

Computation:

Energy used = Power × time

3 bulbs each 60 watt [for 3 hours daily]  = 3 x 60 x 3 x 31 = 16,740 watt = 16.74 kw

4 fans each 100 watt [for 8 hours daily  = 4 x 100 x 8 x 31 = 99,200 watt = 99.2 kw

1 electric heater of 2 kWh [for half hour daily ] = 1 x 2000 x (1/2) x 31 = 31,000 watt = 31 kw

Total Energy consumed = 16.74 + 99.2 + 31

Total Energy consumed = 146.94 kwh

Total cost = 146.94 x 4

Total cost = rs 587.76

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What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?
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The part of the electromagnetic spectrum that can be seen by the human eyes is the visible spectrum. The light waves with the wavelengths of  380 to 740 nm can be sen by the human eyes. Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

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Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

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Part b)

Position of Nth maximum intensity on the screen is given as

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so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

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L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

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Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

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