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trapecia [35]
3 years ago
15

If the total number of earthquakes annually averages 14,500 approximately how many are magnitude 5 or higher? and show work.

Physics
1 answer:
madreJ [45]3 years ago
8 0
  <span>To take a percentage of a number out, simply do the follow: 
Number x Percentage = Answer. 

Therefore: 
14,500x(90/100)=13,050 

To find what percentage a figure represents of a total, do the following simple equation: 
Figure/Total x100 = Percentage. 

So for your Earthquakes, you have a total of: 
Total = 90 + 20 + 5 + 1 = 116 Earthquakes. 

4 to 5 -> 90/116 x100 = 78% to 2 sig. fig. 
5 to 6 -> 20/116 x100 = 17% 
6 to 7 -> 5/116 x100 = 4% 
7+ -> 1/116 x100 = 1% 


</span>The same technique is used everywhere else, sample size divided by total multiplied by 100 gives percentage. Not multiplying by 100 gives you a ratio of 1 that it represents, 1 being 100% (a complete unit). 
<span>Therefore for part 1. of your question I would just do 14,500x0.9=13,050. </span>
<span>You'll probably see that 90/100 = 0.9 since 90% is identical to writing 90/100 or 0.9.</span>
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Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

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