All categories

3 years ago

Where would you find convergent and divergent plate boundaries relative to

Physics

1 answer:

Sergio [31]3 years ago

8 0

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

You might be interested in

You are on a Parkour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal

makvit [3.9K]

Average speed = (total distance covered) / (time to cover the distance)

Total distance covered = (9.5m + 3.5m + 15m) = 28 meters

Time to cover the distance = 43 seconds

Average speed = (28 meters) / (43 seconds)

Average speed = 0.65 meters/second

4 0

3 years ago

A system that can be affected by the outside environment, by an exchange of matter or energy is ______.

larisa86 [58]

$\sf{Answer:}$

A system that can be affected by the outside environment, by an exchange of matter or energy is an open physical system .

4 0

2 years ago

An 8-kg mass is in free fall. What is the velocity of the mass after 9 seconds

zimovet [89]

The velocity of the mass after 9 second is 88 m/s

7 0

3 years ago

Read 2 more answers

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago

Would anyone be kind enough to help me? :)

Travka [436]

Answer:

The racetrack is 996.7 meters long

Explanation:

Convert 251km/h to km/s (kilometers per second)

3600 seconds in an hour, so:

251/3600 = 0.0697km/s

Convert km/s to m/s (meters per second)

1000 meters in a kilometer, so:

0.0697*1000 = 69.7m/s

Find length of racetrack:

69.7m/s*14.3s = 996.7m

If the racer travels 69.7 meters in one second and it takes 14.3 seconds to complete a lap, the racetrack is 996.7 meters long.

6 0

3 years ago