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Monica [59]
3 years ago
12

A well-known transform boundary found in California is the _____. San Francisco fault, San Andreas fault, San Diego fault, OR Sa

n Jose fault?
Physics
2 answers:
Trava [24]3 years ago
7 0
San Andreas fault is the answer
SOVA2 [1]3 years ago
6 0

Answer;

San Andreas fault

A well-known transform boundary found in California is the San Andreas fault.

Explanation;

The San Andreas Fault is the most famous fault in the world.The San Andreas Fault is a place where two tectonic plates touch, the North American and Pacific Plates. The plates are rigid (or almost rigid) slabs of rock that comprise the crust and upper mantle of the Earth. It is about 700 miles long as the crow flies and about 800 miles long when its curves are measured.

-A fault is a planar crack in a rock along which slippage has taken place. Most faults are small; even microscopic; and are not important. Some faults are many miles long.

-Faults can be classified according to which of the three directions of space the rocks on either side move. When the motion is predominantly vertical, they are called dip slip faults. Dip slip faults with dips less than 45 degrees are called thrust faults.

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D) the bending of a wave as it moves around an obstacle or passes through a narrow opening

Explanation:

Even though light travels in straight line, it will bend around objects if their size is comparative to its wavelength. This phenomenon is called diffraction.

Light will also bend if the light travels from one medium to another medium at an angle, but that is called refraction.

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3 years ago
Calculate the accleration of a car if its velocity increases from 15m/s to 75m/s in 5 second​
andrezito [222]

Answer:

I think the acceleration is 12m/s

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Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?
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2 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
At a particular temperature, the speed of sound is 330 m/s and its frequency is 990 Hz. Choose the correct wavelength of this so
koban [17]

Answer:

1/3 M

I hope this helps :) sorry its late

8 0
3 years ago
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