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3 years ago

Along the San Andreas fault in California, tectonic plates are primarily moving _____.

1 answer:

3 0

It's a transform fault because the Pacific Plate and the North American Plate are moving sideways against each other and the fault extends about 800 miles.

Step-by-step explanation:

Occasionally, because the fault is rugged and snaky, the plates get stuck against each other in their movement causing a build-up of energy at the point where they are stuck. When such a point gets unreleased abruptly, the built-up energy is suddenly released causing massive earthquakes. Smaller earthquakes do frequently occur (compared to big ones) as the friction between the two plates causes smaller stick-release points.

Learn More:

For more on the San Andreas fault check out;

brainly.com/question/10975774

brainly.com/question/11681069

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Angle 3 is = to angle 2 because:

Angle 3 & angle 7 are corresponding angles, meaning they have the same measure.

Angle 7 & angle 2 are vertical angles, meaning they also have the same measure.

Therefore, if angle 7 is equal to angle 3, then angle 2 must also be equal to angle 3.

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3 years ago

If a car is moving 30 miles an hour how long will it take to travel 540 miles

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The smallest object visible with your eyes is similar to the width of a piece of hair, which is 1×10−4 meters wide. Using an opt

Sloan [31]

Answer:

B. $5\times10^{2}$

Step-by-step explanation:

We are told that the smallest object visible with our eyes is similar to the width of a piece of hair, which is $1\times 10^{-4}$ meters wide.

Using an optical microscope, we can see items up to $2\times 10^{-7}$ meters wide.

To find the objects we can see with our eyes are how much larger than the objects we can see with an optical microscope, we can set an equation as:

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1*10^{-4}}{2*10^{-7}}$

Using the exponent rule of quotient $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1}{2}*10^{-4-(-7)}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{-4+7}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{3}$

$\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10\times 10^{3-1}$

$\text{The object we can see with our eyes}=5\times10^{2}*\text{The objects we can see with microscope}$

Therefore, the objects we can see with our eyes are $5\times10^{2}$ times larger than the objects we can see with an optical microscope and option B is the correct choice.

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