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3 years ago

What are some of the physical benefits to be derived from aerobic

Physics

1 answer:

VladimirAG [237]3 years ago

6 0

Some of the benefits are increased heart muscles, increase in blood flow, and reduced body fat

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Why is water pollution a concern if water is continuously cycled through Earth’s systems?

kati45 [8]

Answer:Water pollution is a concern because since it is cycled threw the earths system it effect the earths system in a negative way.

Explanation:

I would wait for someone else to answer just in case I’m wrong.

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3 years ago

When the medium is uniform, how do light waves pass through it?

LekaFEV [45]

The correct answer is, A) Straight line motion

I took the quiz

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3 years ago

A 10 N force and a 15 N force are acting from a single point in opposite directions. What additional force must be added to prod

AleksAgata [21]

Answer:

5 N acting in the same direction as the 10 N force

Explanation:

10+5=15

15=15

7 0

3 years ago

Sound source A has a deciBel rating of 50 dB. Sound source B is 1000 times more intense. What is the deciBel rating of B

masha68 [24]

The decibel system of sound intensity operates by a logarithmic scale, meaning that sound intensity increases exponentially in relation to the decibel rating.

For decibels, the equation between intensity and the dB equivalent is:

dB = 10log(i),

where “i” is the intensity of the sound. The ten in front of the log means that an increase in ten dB results in a tenfold increase in sound intensity; for example, a 30 dB sound is ten times softer than a 40 dB sound.

In this case, a sound with a dB of 80 would be 1000 times more intense than a 50 dB sound, so the decibel rating of B is 80.

Hope this helps!

3 0

3 years ago

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago