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velikii [3]
4 years ago
15

Seafloor spreading causes an oceanic plate to collide with a continental plate. If it were to go through the whole tectonic cycl

e, number the steps that would follow.
Physics
2 answers:
Hitman42 [59]4 years ago
7 0
If you're going to ask a question, at least provide the steps or possible answers that you are provided with so that others can help you. Luckily, I know the exact question you are trying to solve. 
The order the steps go in is:
 <span> The oceanic plate would subduct under the continental plate into the asthenosphere. 
From the intense heat and pressure within the mantle, the crust of the oceanic plate would mix with the rock of the mantle. 
Mantle convection would cause magma to rise up through divergent boundaries.</span>
Snowcat [4.5K]4 years ago
4 0

Answer: The entire process is mentioned in  step-wise below-

  1. Firstly, two plates moves opposite to each other, due to this the crust gradually become thin and subsequently rises up.
  2. On continuous thinning of the crust, at one point, the magma will come out to the surface producing a small rift.
  3. Rifting continuously takes place and results in a rift valley,and slowly after millions of years will form an ocean.
  4. Then, the two diverging plate will collide with another oceanic or continental plate, in which the more denser oceanic crust will get subducted beneath the less denser one.
  5. This suducted plate, at much greater depth undergoes partial melting and slowly makes its way up to the crust.
  6. This magma on further pushing towards the surface, initiates sea floor spreading, thus repeating the entire cycle.
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A race car starting from rest accelerates uniformly at a rate of 4.90 meters per squared. What is the cars speed after it has tr
shusha [124]
Ok, we need to find a relation for the speed as it relates to the acceleration.  This is given by the integral of acceleration:

v= \int\limits^{t}_{0} {a} \, dt' =at

Where we have the initial velocity is 0m/s and a will be 4.90m/s².

But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating!  We need to calculate this time by now finding the position function as a function of time.  This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity. 
Position is just the integral of velocity:

s= \int\limits^{}_{} {at} \, dt = \frac{1}{2}at^2

Where the initial velocity and initial position are both zero.

Now we set this position function equal to 200m and find the time, t, it took to get there

\frac{1}{2}(a \frac{m}{s^2} )t^2=200m \\  \\ \frac{1}{2}4.90 \frac{m}{s^2} t^2=200m \\  \\ t^2= \frac{400m}{4.90 \frac{m}{s^2}}=81.63s^2 \\  \\ t= \sqrt{81.63s^2 }  =9.04s

Now let's put t=9.04s into our velocity equation:

v =at=4.9\frac{m}{s^2} \times 9.04s=44.3 \frac{m}{s}


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3 years ago
Can anybody please help? 15 points pls
Mariana [72]
Acceleration= v/ r
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Why the specific heat capacity of the sun remain constant<br><br>​
gayaneshka [121]
The heat remains constant because there’s nothing to cool it down
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3 years ago
A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b
guapka [62]
<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=200m/s is the bullet's initial speed

\theta=0 because we are told the bullet is shot horizontally

t is the time since the bullet is shot until it hits the ground

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=1.5m  is the initial height of the bullet

y=0  is the final height of the bullet (when it finally hits the ground)

g=9.8m/s^{2}  is the acceleration due gravity

<h2>Part (a):</h2>

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}   (3)

0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}   (4)

Finding t:

t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}   (5)

Then we have the time elapsed before the bullet hits the ground:

t=0.553s   (6)

<h2>Part (b):</h2>

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

x=V_{o}cos\theta t   (1)

Substituting the knonw values and the value of t found in (6):

x=200m/s.cos(0)(0.553s)   (7)

x=200m/s(0.553s)   (8)

Finally:

x=110.656m  

4 0
4 years ago
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