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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 200g(m2) mass is pulling the cart (m1) across the table as shown below. If it covers 1.12m in 4.8 seconds starting from rest,

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Answer: 20.21 kg

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$F=m_2g$

⇒F=0.200 kg ×9.8 m/s²=1.96 N

Since there is no other force acting on the cart and there is no friction, so this force will pull the cart.

The Cart covers 1.12 m distance in 4.8 s.

From the equation of motion,

s = u t + 0.5 × a t²

We will find the acceleration of the cart,

Distance covered, s = 1.12 m

Initial velocity, u = 0 ( staring from rest)

Time taken, t = 4.8 s

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Now the force which causes this acceleration is:

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3 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

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Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

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