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2 years ago

11

In the United States, most of our energy comes from this non-renewable resource, which has a negative effect on the environment.

(2 points)
coal
nuclear
natural gas
petroleum

2 answers:

AVprozaik [17]2 years ago

7 0

Answer:

Coal

Explanation:

I took the test.

spin [16.1K]2 years ago

4 0

The best answer among the following choices would be the first option A) Coal.

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Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on

mrs_skeptik [129]

Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

dU = -2.59x10^4 - 6.46^4

dU = -9.05x10^4 kJ

8 0

2 years ago

If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t

Whitepunk [10]

From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer: 283 s (nearest second)

6 0

2 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

2 years ago

Two magnetic poles that repel one another are held three inches apart. What happens when the poles are moved closer together, to

dedylja [7]

Because of the magnets are actually electromagnetics aka what causes them to repel each other the atoms and the electrons will make a force of them pushing away from each other because the two magnetic poles are not north and south

3 0

3 years ago

Read 2 more answers

Potassium-40 has a half-life of approximately 1.25 billion years. Approximately how many years will pass before a sample of pota

Maksim231197 [3]

Answer:

3.75 billion years

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 1.25 billion years

Number of half-lives (n) = 3

Time (t) =?

The time taken for the sample of potassium-40 to contains one-eighth the original amount of parent isotope can be obtained as:

n = t / t½

3 = t / 1.25

Cross multiply

t = 3 × 1.25

t = 3.75 billion years.

Therefore, it will take 3.75 billion years for the sample of potassium-40 to contains one-eighth the original amount of parent isotope

8 0

2 years ago