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2 years ago

What is difference between kilowatt and kilowatt hour?

Physics

2 answers:

Sedaia [141]2 years ago

7 0

Kilowatt hour is a measure of ENERGY
Kilowatt is a measure of POWER

Scrat [10]2 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 rate of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your power 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 rate 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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Answer:

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3 years ago

By what percent does the braking distance of a car decrease, when the speed of the car is reduced by 10.3 percent? Braking dista

likoan [24]

The braking distance is the distance traveled by a car experiencing a braking force until it comes to rest.

Our initial energy is solely kinetic:
$E_i = \frac{1}{2}mv^2$

And, since the car goes to rest, it is no longer in motion. It will have no kinetic energy.
$E_f = 0$

Therefore, there was work done by the braking force.

$W_B = E_f - E_i = -\frac{1}{2}mv^2$

Recall the definition of work:
$W = F\cdot \Delta x$

Or in this case, since the displacement and breaking force are antiparallel:
$W = -F_B\Delta x$

This is equivalent to the dissipation of kinetic energy:
$W = -F_B\Delta x = -\frac{1}{2}mv^2$

Now, to visualize this, let's rearrange the equation to solve for displacement.

$\Delta x =\frac{mv^2}{2F_B}$

There is a direct, SQUARE relationship between necessary braking distance speed.

If the speed was reduced by 10.3 percent, its new speed is only 89.7% percent of the original, so:
$\Delta x' =\frac{m(0.897v)^2}{2F_B}$

$\Delta x' = 0.8046\Delta x$

The reduction by a percentage is:
$1 - 0.8046 = 0.1954 \\\\\boxed{= 19.54\%}$

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1 year ago

A 0.40 kg bead slides on a straight frictionless wire with a velocity of 3.50 cm/s to the right. The

tensa zangetsu [6.8K]

Answer:

Total momentum before collision

P1 =.4 * 3.5 = 1.4 ignoring units here

Total momentum after collision

P2 = .6 * V - .4 * .7 = .6 V - .28

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V = 2.8 cm/sec

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2 years ago

5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.

Aleks04 [339]

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

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3 years ago

Read 2 more answers

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$ , the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q = -3.923\,C$ and $r = 100\,m$ , then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

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3 years ago