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OleMash [197]
3 years ago
15

The US government wants to allocate billions of dollars in the next 10 years to help assure our future energy security. The fund

s will be spread among a variety of possible energy resources. Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas exploration, or a combination of sources? Are there other efforts that should be explored? Support your position with cited information for both questions.
Physics
1 answer:
Tom [10]3 years ago
7 0

This question is a very good one indeed! I am very happy you asked it. I would first like to say that, although energy is important, our planet is more so. That is why I feel like we should always try to explore ideas that are both useful to the human race but also useful in protecting our environment because what is the use in having energy if we will not be able to use it. That is why I feel that we should try at types of energy like solar and wind. The problem is finding ways that are less expensive and more efficient. For example, creating a wind energy system that creates a sizeable about of energy and also is not too expensive and hard to make like the wind turbines of nowadays. Also what comes into question is the effect of these ideas and I feel that although some of these may have harmful effects, they must be used. One thing though is that I do not agree in expanding on our oil and gas exploration because it has so many bad effects to our environment that it is hardly worth it. If we can find alternatives in these other sources, like many states in the USA are trying to do, then we will be able to move much faster in technology and also help protect our lovely planet.

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A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t
dolphi86 [110]
The bicycle is 92% efficient, meaning 92% of energy input is converted to useful (in this case kinetic) energy. Just find 92% (=0.92) of the given input. So:
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8 0
3 years ago
A 15.0 mW laser puts out a narrow beam 2.00 mm indiameter.
Murljashka [212]

Answer:

1341.03 V/m

Explanation:

The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.

                                 S = \frac{P}{A} = cε₀E^{2} _{rms}

                             ⇒ \frac{P}{A} = cε₀E^{2}_{rms}

Where;

P is the power output

A is the area of the beam

c is speed of light

ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m

E_{rms} is the average (rms) value of electric field

Making electricfield E_{rms} the subject of the equation

                                 E^{2}_{rms} = P / Acε₀

                                 E_{rms} = √(P / Acε₀)

But area A = πr²

                                 E_{rms} = √(P / πr²cε₀)                    

Given:

Output power, P = 15 mW = 0. 015 W

Diameter, d = 2 mm = 0.002 m

⇒ Radius, r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m

Solving for average (rms) value of electric field;     

E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }

                                E_{rms} = 1341.03 V/m

                             

                         

                                 

                                 

                                 

6 0
3 years ago
Consider the electric field lines shown in the diagram below. From the diagram, it is apparent that object A is ____ and object
Mariana [72]
Positive charge you gave the lines pointing away, negative charge is pointing toward. Don’t have a photo so I can’t fill in the blanks BUT I can tell you the logic
6 0
2 years ago
What is the acceleration of a car that increases its velocity from 0 to 100 kilometers per hour in 10 seconds?
Lunna [17]

Answer:

0.00278 km/sec2 or 36,000 km/hr2 .

Explanation:

5 0
2 years ago
We measure the loudness of sound in decibels.<br> a. True<br> b. False
Tatiana [17]

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

7 0
3 years ago
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