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Brums [2.3K]
3 years ago
15

If Earth's obliquity was 157 degrees, would the seasons be more severe, less severe, or about the same?

Physics
1 answer:
ikadub [295]3 years ago
7 0
The severity of the seasons on Earth is given not by the distance Earth-Sun but by the tilt of the Earth axis. This happens because that the sun rays are oblique in winter and perpendicular in summer (thus the same quantity of sun rays heats a bigger surface in winter - oblique rays). 
The present tild of the Earth axis is  23.5 degrees (from the vertical). If the axis were tilt at 157 degree this would be equivalent  to 180-157 =23 degree. Thus the severity of the seasons would be approximately the same but the seasons would be reversed (for example instead of winter we would have summer, instead of summer we would have winter). 
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A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
If the potential difference across the bulb in a camping lantern is 9.0 V, what is the
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Answer:

9V

Explanation:

The potential difference across the terminal as the same and thats because we are assuming that the source has no internal resistance.

Internal resistance are usually little resistances in the supply.

4 0
2 years ago
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Answer:

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Explanation:

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KE = ½ mv²

When we triple the velocity, the kinetic energy increases by a factor of 9.

9KE = ½ m(3v)²

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