Tag: work and power

Questions Related to work and power

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$

Show answer
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.

An energy of 4 kJ causes a displacement of 64 m in 2.5 s. The power delivered is

power work and power work, energy and power physics energy and its forms

  1. 16 W

  2. 160 W

  3. 1600 W

  4. 16000 W

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Correct Option: C
Explanation:

Given that


Energy, $E=4\ kJ=4000\ J$
Time taken , $t=2.5\ secs$

Power delivered , $P=\dfrac Et=\dfrac{4000}{2.5}= 1600\ W$

A constant power is delivered to a body moving along a straight line. the distance travelled by the body in time t is proportional to 

power work and power work, energy and power physics energy and its forms

  1. $t ^ { 1 / 2 }$

  2. $t ^ { 3 / 2 }$

  3. $t ^ { 5 / 2 }$

  4. $t ^ { 7 / 2 }$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} power\, P\, =Fv=mav \ acceleration\, a=\frac { S }{ { { t^{ 2 } } } } \, and\, velocity\, v=\frac { S }{ t } , \ where\, \, S\, \, is\, the\, dis\tan  ce\, moved \ Since\, \, the\, power\, P\, and\, mass\, m\, are\, cons\tan  t,\, we\, \, get \ P=\frac { { mS } }{ { { t^{ 2 } } } } \times \frac { S }{ t } \to { S^{ 2 } }={ t^{ 2 } }\to S\propto { t^{ \frac { 3 }{ 2 }  } } \end{array}$

Hence,
option $B$ is correct answer.

In the above question, if the work done on the system along the curved path $ba$ is $52\ J$, heat absorbed is 

power work and power work, energy and power physics energy and its forms

  1. $-140\ J$

  2. $-172\ J$

  3. $140\ J$

  4. $172\ J$

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Correct Option: A

A machine gun fires 360 bullets per minute. Each bullet moves with a  velocity of 600 ms$^{-1}$. If the power of the gun is 5.4 kw, the mass of each bullet is,

power work and power work, energy and power physics energy and its forms

  1. 5 kg

  2. 0.5 kg

  3. 0.05 kg

  4. 5 gm

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Correct Option: D
Explanation:

Given,

Number of bullets, $n=360\,nos.$

Mass of bullet, $m$

Power, $=5400watt$

Kinetic energy of 360 bullets, $K.E=n\times \dfrac{1}{2}m{{v}^{2}}=360\times \dfrac{1}{2}\times m\times {{600}^{2}}=648\times {{10}^{5}}m\,\,J$

$ Power=\dfrac{kineticenergy}{time} $

$ 5400=\dfrac{648\times {{10}^{5}}m}{60} $

$ m=5\times {{10}^{-3\,}}kg\,=\,5\,gram\, $

Hence, weight of each bullet is $5\,gram$

The driving side belt has a tension of $1600\ N$ and the slack side has $500\ N$ tension. The belt turns a pulley $40\ cm$ in radius at a rate of $300\ rpm$. The pulley drives a dynamic having $90\%$ efficiency. How many kilowatt are being delivered by the dynamo?

power work and power work, energy and power physics energy and its forms

  1. $12.4\ kW$

  2. $6.2\ kW$

  3. $24.8\ kW$

  4. $13.77\ kW$

Show answer
Correct Option: A