10 Points

What are some ways to conserve electricity

emmasim [6.3K]

Turn off lights when leaving rooms.

Unplug unused appliances. Even when powered off these appliances use electricity.

Replace regular light bulbs with energy saving bulbs.

7 0

3 years ago

Read 2 more answers

A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pus

Alchen [17]

Answer:

<h3>2,321.62Joules</h3>

Explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone = 2,321.62Joules

<em>Hence the amount of work done is 2,321.62Joules</em>

3 0

3 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

The boiling point of sulfur is 444.6 celsius .sulfur melting point is 586.1 fahrenheit lower than its boiling?

iVinArrow [24]

Answer:

(a) Melting point is 136.8°C

(b) Melting point is 278.24°F

Boiling point is 832.28°F

(c) Melting point is 409.8K

Boiling point is 717.6K

Explanation:

(a) 586.1°F = 5/9(586.1 - 32)°C = 307.8°C

Melting point = 444.6°C - 307.8°C = 136.8°C

(b) Melting point = 136.8°C = (9/5×136.8) + 32 = 278.24°F

Boiling point = 444.6°C = (9/5×444.6) + 32 = 832.28°F

(c) Melting point = 136.8°C = 136.8 + 273 = 409.8K

Boiling point = 444.6°C = 444.6 + 273 = 717.6K

8 0

3 years ago

What is the difference between amplitude and wavelength?

jasenka [17]

Answer:

Letter b is wavelength. Letter a is amplitude.

Explanation:

Let's imagine a simple experiment. Imagine you have a long thick rope which one end is at your hands, and you start an oscillatory motion in it, moving your hand up and down. Then a friend of you take a picture of the rope in motion, looking at the rope laterally. Now let's find the wavelength and amplitude. Amplitude is "The distance from the center of the oscillation of the rope (when the rope was not in motion) to its high or low point", or the vertical displacement, in our experiment. On the other hand, wavelength is "The distance between one high point /low point and the next high point /low point". Take a look at a photo of a wave in your textbook and you will find the answer as well. ; )

3 0

3 years ago