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2 years ago

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

1 answer:

8 0

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%

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).
Therefore for part 1. of your question I would just do 14,500x0.9=13,050. 
You'll probably see that 90/100 = 0.9 since 90% is identical to writing 90/100 or 0.9.

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2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

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The value is $v = 2.3359 *10^{4} \ m/s$

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From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

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$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

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=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

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=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

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