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Alekssandra [29.7K]
3 years ago
10

What is difference between kilowatt and kilowatt hour?

Physics
2 answers:
Sedaia [141]3 years ago
7 0
Kilowatt hour is a measure of ENERGY
Kilowatt is a measure of POWER
Scrat [10]3 years ago
6 0
Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate.  The other is a quantity, after maintaining a rate for some time.

-- 'Joule' is a unit of energy.  It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.

-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.

-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.

-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.

-- 'Kilowatt' is a bigger <em>rate</em> of using energy . . . 1,000 joules per second.

-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .

-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .

Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.


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An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A
melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

\omega = \frac{\theta}{t}

<em>Where θ: is the angular displacement = π (since it moves halfway through a complete circle)</em>

We have:

t = \frac{\theta}{\omega} = \frac{\theta}{v/r}  

<u>Where</u>:      

<em>v: is the tangential speed </em>

<em>r: is the radius</em>

The radius can be found equaling the magnetic force with the centripetal force:

qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!    

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3 years ago
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