Tag: commercial unit of energy

Questions Related to commercial unit of energy

1 kWh is equal to

physics work and energy commercial unit of energy power work and power

  1. $3.6 \times 10^6 MJ$

  2. $3.6 \times 10^5 MJ$

  3. $3.6 \times 10^2 MJ$

  4. $3.6 MJ$

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Correct Option: D
Explanation:
1 kilowatt hour is the energy produced by 1 kilowatt  power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

$\implies 1kWh=3.6\times 10^6J$

$\implies 1kWh=3.6MJ$

Answer-(D)

Number of kilowatt-hours =$\dfrac { volt\times ampere\times time }{ 1000 } $. Then:

physics work and energy commercial unit of energy power work and power

  1. time in seconds

  2. time in minutes

  3. time in hours

  4. time in days

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Correct Option: C
Explanation:

Kilowatt-hours is the power generated in one hour=$\dfrac{volt\times current\times time( hour)}{1000}$


Answer-(C)

If 1 unit of electricity cost $0.20$, how much does it cost to switch on a heater marked $120 V$, $3 A$ for $90$ min?

physics work and energy commercial unit of energy power work and power

  1. $ 0.11$

  2. $ 2.70$

  3. $ 64.80$

  4. $ 108.00$

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Correct Option: A
Explanation:
Voltage across the heater $V = .12$ kilo-volts 
Current flowing through the heater $I = 3 A$
Thus power of the heater $P = VI$
$\therefore$ $P = 0.12 \times 3 = 0.36 $ $kW$
Time of operation $t = 90$ min $ = 1.5 $ hr
Thus energy consumed $E = Pt$
$\implies$ $E = 0.36 \times 1.5 = 0.54$ $kWhr$
Cost to switch on heater =  $0.54 \times 0.2 = 0.11$

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P _{0}$. The instantaneous velocity of this car is proportional to:

physics work and power commercial unit of energy power work and energy

  1. $t^{1/2}$

  2. $t^{-1/2}$

  3. $t/\sqrt{m}$

  4. $t^{2}P _{0}$

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Correct Option: A
Explanation:

Power =F.V


Power delivered as at 

Cont. magnitude $P _0$

$P _0=F.V$

$P _0=ma\times V$

$\dfrac{P _0}{m}=\dfrac{dv}{dt}\times V$

$\displaystyle \left(\dfrac{P _0}{m}\right)\int^t _0 dt=\int^v _0 vdv$

$\dfrac{P _0t}{m}=\dfrac{V^2}{2}$

$V^2=\left(\dfrac{2P _0}{m}\right)t$

$V=\sqrt{\dfrac{2P _0}{m}}\times t^{\dfrac{1}{2}}$

$\boxed{V\alpha\,t^{\dfrac{1}{2}}}$

$1$ kWh$=$ ______________J.

physics work and energy commercial unit of energy power work and power

  1. $3.6\times 10^6$

  2. $36\times 10^6$

  3. $3.6\times 10^7$

  4. $3.6\times 10^5$

Show answer
Correct Option: A
Explanation:

$1$ kilowatt-hour(kWh) is a unit of energy. Normally, we want energy to be expressed in joules(J) and time in seconds(s).
Energy(kWh)$=$Power(kW)$\times$(h)$=1000$W$\times 3600$s$=1000$J/s$\times 3600$s$=3600000$J$=3.6\times 10^6$J.