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3 years ago

What do you notice about the lines of earthquakes and volcanoes and where they seem to occur on Earth's surface?

Physics

1 answer:

Stolb23 [73]3 years ago

3 0

Answer:

Most earthquakes and volcanoes occur because of the movement of the plates, especially as plates interact at their edges or boundaries.

Explanation:

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2 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

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3 years ago

Becky leaves home and rides a distance of 30km it took her 2.5 hours what is her speed

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Her speed would be 12.5km p/s

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