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3 years ago

Which explanations provide support for continental drift theory? Check all that apply.

Physics

1 answer:

denis23 [38]3 years ago

8 0

<h3><u>Full Question:</u></h3>

Which explanations provide support for continental drift theory? Check all that apply.

Dinosaurs lived on many continents.

Coal fields match up across continents.

Tropical plant fossils were found in Arctic areas.

Similar rock types are found across continents.

Evidence of glaciers can be found in South Africa.

Coal fields match up across continents.

Similar rock types are found across continents.

Tropical plant fossils were found in Arctic areas.

Evidence of glaciers can be found in South Africa.

<h3><u>Explanation:</u></h3>

The way the continents on the earth shifts its position is explained by the Continental drift theory. Alfred Wegener was the one who fist proposed this theory in 1912. This theory also explains how similar rocks are formed on the different continents and also the reason why some of the animal and plant fossils looks similar.

The evidences that supports the continental drift theory can be Coal fields match up across continents. Fossils of Glossopteris supports the theory of continental drift with coal fields and coastlines . The land features, climate changes are some of the evidences that also supports the theory of continental drift.The evidences of the glaciers that can be found in South Africa also supports the explanation of the continental drift.

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A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00

Airida [17]

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

7 0

2 years ago

Ian’s school is exactly 6 blocks north of his house. What is his average velocity on a walk from home to school that takes him 1

choli [55]

Answer:

.5 units north

55 ft/minutes(<em>squared</em>) north

Explanation:

.5 is what your info gives me but if i take that the average distance a block is it is 660ft than the answer is 55.

4 0

3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$