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Black_prince [1.1K]
3 years ago
13

What is likely happening at the boundary between the Nazca plate and the Pacific plate?

Physics
2 answers:
Nataly_w [17]3 years ago
8 0
<span>The eastern margin is a convergent boundary subduction zone under the South American Plate and the Andes Mountains, forming the Peru–Chile Trench. The southern side is a divergent boundary with the Antarctic Plate, the Chile Rise, where seafloor spreading permits magma to rise.</span>
Alika [10]3 years ago
8 0

Explanation:

The south-eastern side of the Pacific plate is a divergent boundary (the plates are pulled, and not pushed, apart) with the Nazca Plate forming the East Pacific Rise.

The East Pacific Rise is part of the circumglobal system of active volcanic ridges, all of which define the position of diverging plates where new crust is being created. These ridges represent the locus of global shallow-water earthquakes.

I hope you find this information useful and interesting! Good luck!

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The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The
IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

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3 years ago
Cool facts about satellites ???
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Answer:

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Two major characteristics that change when air is heated and cooled is the density and the direction of the air flow. They change the movement of the air by heated air rises and cooled air doesn't. 
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