Energy Units and Forms - Class IX

Given,

joule and calorie are both units of energy .
$1\  joule = 0.24 \ calories$

We know that $ power = work/ time$.

Energy consumed in a month
$=250 : units = 250 : kWh$
$= 250 \times 3.6 \times 10^6 = 9 \times 10^8:J $

The kilowatt-hour ($kWh$) is a unit of energy equivalent to one kilowatt ($1\  kW$) of power expended for one hour ($1\  h$) of time.

False
because,The kilowatt-hour (kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour (1 h) of time.

The commercial unit of measurement energy is Kilowatt hour.

$1000\ W= 1\ kW$
$1000 \ kW= 1\ MW$

This is unit of energy.If the energy is being transmitted or used at a constant rate over a period of time, the total energy in kilowatt hours is the power in kilowatts multiplied by the time in hours.

Joule, kilowatt Hour and Ergs are different units of the physical quantity Energy.

Power, $P=VI$

Since, we know that SI unit of energy is represented by Joule. Hence, let us convert all the different units in the form of Joules. 

As, $1kW=1000W ; 1hr=3600s$
$1kWh=1000Wh=3600000=3.6\times10^6Ws=3.6\times10^6J$

We know that the electrical potential energy of electron $U=eV$ where e is the charge of electron and V be the voltage. So electron volt will be the unit of energy.  

Unit of energy is $ML^2T^{-2}$ 

1 horse power is equal to 746 watts.

1 watt is defined as 1 joule/second.

So $1kW = 1000W$, and $1 hour =60 \times 60 seconds $

So $1kWh = 3600kWs = 3,600,000Ws = 3,600,000J$

$1kWh=3.6*{{10}^{6}}J$

Watt is defined as joule per second. Kilowatt is $10^3$joule per second and kwh is kilowatt hour $10^3\times60\times60$ 

1 kWh is equal to 3.6 $\times$ 10$^{6}$

The kilowatt hour is a unit of energy equal to $3.6\times 10^6J$. If the energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt hours is the power in kilowatts multiplied by the time in hours. so if kWh is unit of energy so none of the option is correct hence option D is the answer.

A watt second (also watt-second, symbol W s or W. · s) is a derived unit of energy equivalent to the joule. The watt-second is the energy equivalent to the power of one watt sustained for one second.

calorie (cal) is the energy needed to increase 1 gram of water by 1°C at a pressure of 1 atmosphere.

 Electron volt (symbol eV) is a unit of energy equal to approximately 1.6×10⁻¹⁹  joules (symbol J). By definition, it is the amount of energy gained (or lost) by the charge of a single electron moving across an electric potential difference of one volt.

no chemical reaction takes place while wounding a spring so no chemical energy.

An erg is the amount of work done by a force of one dyne exerted for a distance of one centimeter. In the CGS base units, it is equal to one gram centimeter-squared per second-squared

A flying helicopter has potential energy,mgh(mass by g by weight) and this potential energy does not depend in any way on how fast it is flying in a horizontal direction.

One kilowatt-hour is the energy consumed by a 1 kW device in 1 hour. Since 1 kW = 1000 J/s and 1 hour = 3600 s, 1 kWh = 1000 * 3600 = 3.6 * 10^6 Joules.

In presence of light energy, other objects are seen.

Charge is a fundamental property of matter. It is not a form of energy.

A compressed spring stores elastic potential energy. When released, this potential energy is converted into kinetic energy as the spring moves.

The negative of the work done  by the conservative internal forces on a system equals to the change in potential energy

Horsepower (hp) is a unit of measurement of power (the rate at which work is done).A power level of 1 hp is approximately equivalent to $746$ watt (W) or $0.746$ kilowatt (kW).

In the SI system, 1 Joule = 1 kg * m^2 / s^2. In the CGS system, 1 erg = 1 g * cm^2 / s^2. Since 1 kg = 10^3 g and 1 m^2 = 10^4 cm^2, 1 J = 10^3 * 10^4 = 10^7 ergs.

The Watt is a unit of power, defined as joules per second, not a unit of energy. Electron-volt, Joule, and Calorie are all standard units used to measure energy.

One unit of electricity is defined as one kilowatt-hour (kWh). As calculated previously, 1 kWh = 3.6 * 10^6 Joules.

By the work-energy theorem, the total work done equals the change in kinetic energy. The work done by the external force is -10 J. The change in kinetic energy is 1/2 * m * (v_B^2 - v_A^2) = 1/2 * 2 * (4 - 16) = -12 J. The change in potential energy is the negative of the work done by the conservative (electrostatic) force. Given the total work W_ext + W_elec = Delta K, we find Delta U = +10 J.

$1\ kcal=4.2\times1000=4.2\times10^3\ J$

Energy required to accelerate from $\displaystyle 10:m/s$ to $20:m/s$ $E _{1}=\frac{1}{2}m(400-100)$

The electrical unit for the electricity of home is used as kilowatt hour (kWh). Here the cost of 1 kWh electricity is Rs 150.  

This is unit of energy $ML^2T^{-2}$ (in dimensional form).

The kilowatt hour is a unit of energy equal to 3.6 megajoules. If the energy is being transmitted or used at a constant rate over a period of time, the total energy in kilowatt hours is the power in kilowatts multiplied by the time in hours.

the given unit $kgm^2s^{-2}$ is unit of energy and  any of the energy either potential or kinetic will have  same unit 

In this problem, we have to convert the heat energy in unit from calorie to Joule. 

$1Wh=1W\times{1h}$

$=1{J}.{s}^{-1}\times{3600}s$

$=3600J=3.6kJ$

Electrical energy $V \times l \times t$
Electric energy unit $= kW  hour$

$1kWh=1kW\times{1h}$

$=1000{J}.{s}^{-1}\times{3600}s$

$=3.6\times{10}^{6}J=3.6MJ$

Intensity = 60 dB

Mechanical energy & potential energy can be negative.

Unit of Energy = Unit of Force $\times$ Unit of length 
So if we increase unit of length and force, each by four times, then unit of energy will increase by sixteen times.