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3 years ago

5

Geologists map out ________ from previous earth quakes to determine earthquakes risk. a. graphs c. data b. buildings d. none of

the above
Took the test, it is C

1 answer:

photoshop1234 [79]3 years ago

8 0

Answer: Geologists map out <em>data </em>from previous earth quakes to determine earthquakes risk.

Explanation:

Geology is the study of earth. Also, it is the study of how Earth's materials, structures etc are changed over time.

Geologist is a scientist who studies the structure of earth, rocks, soil, fossils and earthquake. So, they map out the data from previous earthquakes to determine earthquake risk.

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A 375-pound concrete cylinder has a base area of 144 square inches. with the cylinder resting on its base, the pressure exerted

Akimi4 [234]

The pressure exerted by the concrete cylinder is 2.60 pound/in².

We need to know about the pressure to solve this problem. Pressure is a unit that describes how much force is applied to a surface area. It can be determined as

P = F / A

where P is pressure, F is force and A is area.

From the question above, we know that

F = 375 pound

A = 144 in²

By substituting the given parameters, we can calculate the pressure

P = F / A

P = 375 / 144

P = 2.60 pound/in²

Thus, the pressure should be 2.60 pound/in².

Find more on pressure at: brainly.com/question/25965960

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6 0

1 year ago

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m

denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0

2 years ago

A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well

Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

Explanation:

8 0

2 years ago

A player kick the soccer ball from ground level and send it flying at an angle of 30° at a speed of 26M/S. What is the maximum h

icang [17]

The answer would be 2.63. Your welcome. This has been changed to the correct answer.

7 0

3 years ago

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

2 years ago