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castortr0y [4]
3 years ago
15

What determines how the plates interact at their boundaries

Physics
1 answer:
swat323 years ago
3 0

Answer:

Tectonic plate interactions are of three different basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.

Explanation:

Meaning the answer to your question is depending on what type of tectonic plate interaction is occurring will depend on how the plates interact.

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A 30 kg child went down a 10 m tall slide. Assuming no energy was lost as friction, what was the child's velocity when he reache
AveGali [126]
14 m/s or 50km/h. See the details in the attached picture.

5 0
3 years ago
If you weigh 882 N on EARTH (HINT: What number do we ALWAYS use for gravity on Earth), what is your mass? ​
Softa [21]
882 divided by 9.81 (this is acceleration due to gravity) it equals 89.91
4 0
3 years ago
If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1
Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

I=\frac{Q}{t}

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
3 0
1 year ago
A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol
il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

  • When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
  • These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
  • In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
  • Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
  • This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
  • At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

       F_{c} = N + m*g (1)

  • Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

       F_{c} = m*\frac{v^{2}}{r} (2)

  • Since the normal force takes any value as needed to make (1) equal to (2),  if the speed diminishes, it will be needed less force to keep the equality valid.
  • In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
  • In this condition, from (1) and (2), we can find the minimum possible value of  the speed that still keeps the motorcycle touching the surface, as follows:
  • v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)
6 0
3 years ago
You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
Monica [59]

a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.

from above statement we got

height = 78.4 m

since the ball is thrown, so its vertical velocity would be zero

u = 0

taking g = 9.8m/s^2

now, using the equation of motion

h = ut + gt^2/2

now putting all the values in it

we got ,

78.4 = 9.8 * t^2/ 2

by solving we got,

t = 4 sec

b) now, since along the horizontal , no force acting and accelaration is zero so

R = ut , R is RANGE

R = 5 * 4

range =  20 m

c)  vertical components of the stone’s velocity just before it hits the ground = v sin θ =

horizontal   components of the stone’s velocity  just before it hits the ground = v cos θ

To know more about velocity  visit :

brainly.com/question/18084516

#SPJ9

4 0
1 year ago
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