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7nadin3 [17]
2 years ago
11

Lightbulb A is marked "25 W 120V, "and lightbulb B is marked "100 W 120V " These labels mean that each lightbulb has its respect

ive power delivered to it when it is connected to a constant 120-V source.(f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at 0.110 per kWh.
Physics
1 answer:
Dovator [93]2 years ago
3 0

The cost of running the lightbulb A for 30 days at 0.110 per KWh is 1.98

<h3>How to determine the energy </h3>

We'll beging by calculating the energy used by lightbulb A. This can be obtained as follow:

  • Power (P) = 25 watts = 25 / 1000 = 0.025 KW
  • Time (t) = 30 days = 30 × 24 = 720 h
  • Energy (E) =?

E = Pt

E = 0.025 × 720

E = 18 KWh

<h3>How to determine the cost for running the bulb for 30 days</h3>

The cost of running the bulb for 30 days can be obtained as follow:

  • Cost per KWh = 0.11
  • Energy (E) = 18 KWh
  • Cost =?

Cost = energy × Cost per KWh

Cost = 18 × 0.11

Cost = 1.98

Lean more about buying electrical energy:

brainly.com/question/16963941

#SPJ4

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\theta = 67.22 degree

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now from above two equations

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7 0
2 years ago
A 2.98 nF parallel-plate capacitor is charged to an initial potential difference of 49 V and then isolated. The dielectric mater
I am Lyosha [343]

Answer:

Potential difference will be 151.9 volt  

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We have given capacitance of the capacitor C=2.98nF=2.98\times 10^{-9}F

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We know that when a dielectric medium is introduced then p[otential difference is increases by k times

As the dielectric constant k = 3.1

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8 0
3 years ago
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Part a)

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Part b)

maximum speed = 0.58 m/s

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As we know that angular frequency of spring block system is given as

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here we know

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Part b)

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