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3 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

<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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Assume that a person bouncing a ball represents a closed system. Which statement best describes how the amounts of the ball's po

Zinaida [17]

Answer:

Explanation:

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3 years ago

The key differences between rotational kinematics and translational kinematics is: A. Rotational kinematics must specify an axis

inn [45]

In Translational motion, the body moves in a straight line whereas in Rotational motion, the body moves in a circular path.

<h3>Differences between rotational kinematics and translational kinematics</h3>

In Translational motion, the body moves in a straight line and the body same displacement in equal interval of time while on the other hand, in Rotational motion, the body moves in a circular path and the body travels same angular displacement in equal interval of time..

Learn more about motion here: brainly.com/question/453639

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2 years ago

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at

vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

$g=-9.8 m/s^2$ (acceleration of gravity)

The initial vertical velocity is

$u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s$

where the negative sign means the direction is downward.

The vertical position of the ball is given by

$y(t) = h + u_y t + \frac{1}{2}gt^2$

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

$0 = 4.50 -1.96t-4.9t^2$

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

$v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s$

where u = 3.05 m/s is the initial speed and $\theta$ the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

$d=v_x t$

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

$d=(2.34)(0.78)=1.83 m$

7 0

3 years ago

Is the classification for an instrument that produces sound whne a string or strings stretched between two points is plucked?

Yuki888 [10]

The correct answer for the question is Chordophone
Chordophone is an instrument in which a stretched, vibrating string produces the initial sound. Strings instruments produce sound through the vibration of strings. The length, tightedness, and thickness determines the sound produced by the strings.

5 0

3 years ago

An 81.5-kg man stands on a horizontal surface.

Elza [17]

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10² N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

$P=\dfrac{F}{A}$

$P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2$

P=181.11 x 10² N/m²

4 0

3 years ago