The Socio-economic Activites of Australia

Australia has unique plant and animal life. Western Australia produces more than the half a number of metallic minerals. This note has information about Australia's economic and social activities.

Summary

Australia has unique plant and animal life. Western Australia produces more than the half a number of metallic minerals. This note has information about Australia's economic and social activities.

Things to Remember

  • Australia has unique plant and animal life.
  • Australia is very rich in minerals.
  • Western Australia produces more than the half the amount of metallic minerals.
  • Great Barrier Reef, Sydney, Brisbane beaches, vast deserts and historic sites all allure tourists from throughout the world.
  • Sugarcane is cultivated especially in the coast of Queensland.

MCQs

No MCQs found.

Subjective Questions

Q1:

Determine whether the following quantities are vectors or scalars. Give reasons for your answers.
1. Force                  2. Temperature
3. Length            
4. Displacement      5. Velocity
6. Acceleration       7. Mass
8. Energy              9. Time

 


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>1. Force is a vector quantity because it has both magnitude and direction. <br />2. Temperature is a scalar quantity because it has magnitude only. <br />3. Length is a scalar quantity because it has magnitude only. &nbsp;<br />4. Displacement is a vector quantity because it has both magnitude and direction. <br />5. Velocity is a vector quantity because it has both magnitude and direction. <br />6. Acceleration is a vector quantity because it has both magnitude and direction.<br />7. Mass is a scalar quantity because it has magnitude only. <br />8. Energy is a scalar quantity because it has magnitude only.<br />9. Time is a scalar quantity because it has magnitude only.</p>

Q2:

Express the following vector  joining the points A and B vector \(\overrightarrow{AB}\) in th form of column vectors and find the magnitude and direction of  \(\overrightarrow{AB}\) .

A (1 , -1) , B (2 , 4)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given points A (1, -1) , B(2 , 4)</p>
<figure class="inline-right" style="width: 300px;"><img src="/uploads/225_b.png" alt="figure" width="258" height="290" /><figcaption>figure</figcaption></figure>
<p><br />\(\therefore\) &nbsp;\(\overrightarrow{AB}\) =\(\begin{pmatrix} 2-1 \\ 4+1 \\ \end{pmatrix}\)&nbsp;</p>
<p>=&nbsp;\(\begin{pmatrix} 1 \\ 5&nbsp;\\ \end{pmatrix}\)&nbsp;<br /><br />Magnitude of vector | \(\overrightarrow{AB}\) | = \(\sqrt{1^{2} + 5^{2}}\)<br /> = \(\sqrt{1 + 25}\)<br /> = \(\sqrt{26}\) Ans. <br /><br />Direction of \(\overrightarrow{AB}\) is tan \(\theta\) = \(\frac{5}{1}\)= tan(tan<sup>-1</sup>5) <br />\(\therefore\) \(\theta\) = tan<sup>-1</sup>5</p>

Q3:

Express the following vectors joining the points A and B (vector \(\overrightarrow {AB}\) in the form of column vectors and find the magnitude and direction of \(\overrightarrow {AB}\) .
A(3 , -4)  , B (-2 , 1)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given points A(3 , -4) , B(-2 , 1)</p>
<figure class="inline-right" style="width: 300px;">
<figure class="inline-right" style="width: 300px;"><img src="/uploads/225_c1.png" alt="1" width="300" height="339" /> <figcaption><br /></figcaption></figure>
<br /><figcaption>figure</figcaption></figure>
<p><br />\(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} -5 \\ 5 \end{pmatrix}\) Ans. <br /><br /><br />Magnitude of \(\overrightarrow {AB}\) = \(\sqrt { (-5) ^{2} + 5^{2}}\)<br /> = \(\sqrt{25 + 25}\)<br /> = 5\(\sqrt{2}\) units. <br />Direction of \(\overrightarrow {AB}\) , <br />tan \(\theta\) = \(\frac{5}{-5}\) = -1 = tan135<sup>o. </sup>Ans.</p>

Q4:

Express the following vectors joining the points A and B (vector \(\overrightarrow {AB}\) in the form of column vectors and find the magnitude and direction of \(\overrightarrow {AB}\) .
A(1 , 1) , B(2 , 4)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given points A(1 , 1) , B(2 , 4)</p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/225_d.png" alt="figure" width="250" height="282"><figcaption>figure</figcaption></figure><p><br>\(\therefore\) Vector \(\overrightarrow {AB}\) = \(\begin{pmatrix}1 \\ 3 \end{pmatrix}\) <br>Magnitude of \(\overrightarrow {AB}\) = \(\sqrt{1 + 3^{2}}\)<br> = \(\sqrt{ 1+9}\)<br> = \(\sqrt{10}\) units. Ans. <br><br>Direction of \(\overrightarrow {AB}\) is tan \(\theta\) = \(\frac{3}{1}\) = tan(tan<sup>-1</sup>3)<br>\(\theta\) = tan<sup>-1</sup>3 Ans.</p>

Q5:

Express the following vectors joining the points A and B (vector \(\overrightarrow {AB}\) in the form of column vectors and find the magnitude and direction of \(\overrightarrow {AB}\) .
A(-1 , 3)  , B(1 , -3)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given points A( -1 , 3) , B(1 , -3)</p>
<figure class="inline-right" style="width: 300px;">
<figure class="inline-right" style="width: 300px;"><img src="/uploads/e3.png" alt="1" width="300" height="296" /> <figcaption><br /></figcaption></figure>
<br /><figcaption><br /></figcaption></figure>
<p><br />\(\therefore\) Vector \(\overrightarrow {AB}\) = \(\begin{pmatrix} 2 \\ -6 \end {pmatrix}\) Ans. <br /><br />\(\therefore\) Magnitude of \(\overrightarrow {AB}\) = \(\sqrt{2 ^{2} + 6 ^{2}}\)<br /> = \(\sqrt{4 + 36}\)<br /> = \(\sqrt{4 + 36}\)<br /> = \(\sqrt{40}\)<br /> = 2\(\sqrt{10}\)units. Ans. <br /><br />Direction of \(\overrightarrow {AB}\) , tan\(\theta\) = \(\frac{-6}{2}\)<br /> = -3<br /> = tan(tan<sup>-1</sup>(-3)</p>
<p>\(\therefore\) \(\theta\) =tan<sup>-1</sup>(-3) .</p>

Q6:

Express the following vectors joining the points A and B (vector \(\overrightarrow {AB}\) in the form of column vectors and find the magnitude and direction of \(\overrightarrow {AB}\).
A( -2 , 0) , B(-3 , -2)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Given points A(-2 , 0) , B(-3 , -2)</p>
<figure class="inline-right" style="width: 250px;">
<figure class="inline-right" style="width: 300px;"><img src="/uploads/saDazvydbhntyd1.png" alt="1" width="300" height="296" /> <figcaption><br /></figcaption></figure>
<br /><figcaption><br /></figcaption></figure>
<p><br />\(\therefore\) Vector \(\overrightarrow {AB}\) = \(\begin {pmatrix} - 1 \\ -2 \end{pmatrix}\) Ans. <br /><br />Magnitude of \(\overrightarrow {AB}\) = \(\sqrt { -1^{2} + -2 ^{2}}\)<br /> = \(\sqrt{1 +4}\)<br /> = \(\sqrt{5}\) Ans. <br /><br />Direction of \(\overrightarrow {AB}\) , tan\(\theta\) = \(\frac{-2}{-1}\)<br /> = tan(tan<sup>-1</sup>2)<br /> \(\theta\) = (tan<sup>-1</sup>2)</p>

Q7:

For every vector \(\overrightarrow{AB}\) , find the negative vectors - \(\overrightarrow{AB}\) or \(\overrightarrow {BA}\) and then express them in then form of column vectors.
A (0 , 0) , B(1 , 4)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , A(x<sub>1</sub> y<sub>1</sub>) = (0 , 0) , B = (1 , 4)<br />Now , x -component of \(\overrightarrow {AB}\) = x<sub>2</sub> - x<sub>1</sub> = 1 - 0 = 1<br /> y- component of \(\overrightarrow {AB}\) = y<sub>2</sub> - y<sub>1</sub> = 4 - 0 = 4<br /><br />\(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} 1 \\ 4 \end {pmatrix}\)<br />\(\therefore\) - \(\overrightarrow {AB}\) = \(\overrightarrow {BA}\) = -1 \(\begin {pmatrix} 1 \\ 4 \end {pmatrix}\) = \(\begin {pmatrix} -1 \\ -4 \end {pmatrix}\) Ans.</p>

Q8:

For every vector \(\overrightarrow {AB}\) , find the negative yectors - \(\overrightarrow {AB}\) or \(\overrightarrow {BA}\) and then express them in then form of column vectors.
A(1 , -1) , B(2 , 4)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , A(x<sub>1</sub> , y<sub>1</sub>) = (1 , -1) , B(x<sub>2</sub> , y<sub>2</sub>) = (2 , 4)<br />Now , x - component of \(\overrightarrow {AB}\) = x<sub>2</sub> - x<sub>1</sub> = 2 - 1 = 1<br /> y - component of \(\overrightarrow {AB}\) = y<sub>2</sub> - y<sub>1</sub> = 4 - (-1) = 4 + 1 = 5. <br /> \(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} 1 \\ 5 \end{pmatrix}\) <br /> \(\therefore\) - \(\overrightarrow {AB}\) = \(\overrightarrow {BA}\) = -\(\begin {pmatrix} 1 \\ 5 \end{pmatrix}\) = \(\begin {pmatrix} -1 \\ -5 \end{pmatrix}\) Ans.</p>

Q9:

For every vector \(\overrightarrow {AB}\) , find the negative yectors - \(\overrightarrow {AB}\) or \(\overrightarrow {BA}\) and then express them in then form of column vectors.

A(-1 , 3) ,B(1 , -3)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , A(x<sub>1</sub> , y<sub>1</sub>) = (-1 , 3) , B(x<sub>2</sub> , y<sub>2</sub>) = (1 , -3)<br />Now , x - component of \(\overrightarrow {AB}\) = x<sub>2</sub> - x<sub>1</sub> = 1 -(- 1) =&nbsp;2<br />\(\therefore\) y - component of \(\overrightarrow {AB}\) = y<sub>2</sub> - y<sub>1</sub> = -3 - 3 = -6<br />\(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} 2 \\ -6 \end{pmatrix}\) <br />\(\therefore\) - \(\overrightarrow {AB}\) =\(\overrightarrow {BA}\) = - \(\begin {pmatrix} 2 \\ -6 \end{pmatrix}\) = \(\begin {pmatrix} -2 \\6 \end{pmatrix}\) Ans.</p>

Q10:

For every vector \(\overrightarrow {AB}\) , find the negative vector - \(\overrightarrow {AB}\) or \(\overrightarrow {BA}\) and then express them in the form of column vectors.
A(2 , 0) , B(-3 , -2)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , A(x1 , y1) = (2 , 0) , B(x2 - x1) = (-3 , -2)</p> <p>Now , x - component of \(\overrightarrow {AB}\) = x2 - x1 = -3 + 2 = -1<br> y - component of \(\overrightarrow {AB}\) =y2- y1 = -2 ,-0 -2<br>\(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} -1 \\ -2 \end{pmatrix}\) <br>\(\therefore\) -\(\overrightarrow {AB}\) = \(\overrightarrow {BA}\) = \(\begin{pmatrix} -1 \\ -2 \end{pmatrix}\) = \(\begin {pmatrix} 1 \\ 2 \end {pmatrix}\) Ans.</p>

Q11:

If the position vector of A is \(\overrightarrow {OA}\) = \(\begin {pmatrix} 3 \\ 4 \end{pmatrix}\) and the position vector of B is \(\overrightarrow {OB}\) = \(\begin {pmatrix} 1 \\ 5 \end {pmatrix}\) , find the magnitude and direction of \(\overrightarrow {AB}\).


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here given , <br />Position vector of A is \(\overrightarrow {OA}\) = \(\begin {pmatrix} 3 \\ 4 \end{pmatrix}\)<br />\(\therefore\) A(x<sub>1</sub> , y<sub>1</sub>) = (3 , 4)<br />Position vector of B is \(\overrightarrow {OA}\) = \(\begin {pmatrix} 1 \\ 5 \end{pmatrix}\) <br />\(\therefore\) B (x<sub>2</sub> , y<sub>2</sub>) = (1 , 5)<br />Now , x - component of \(\overrightarrow {AB}\) = x<sub>2</sub> - x<sub>1</sub> = (1 -3) = -2<br /> y - component of \(\overrightarrow {AB}\) = y<sub>2</sub> - y<sub>1</sub> = 5 - 4 = 1<br />\(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} -2 \\ 1 \end{pmatrix}\) <br />\(\therefore\) Magnitude of \(\overrightarrow {AB}\) = \(\sqrt {x -component ^{2} + y - component ^{2}}\)<br />= \(\sqrt{(-2) ^{2} + 1^{2}}\) = \(\sqrt{(4 + 1)}\) = \(\sqrt{5}\) unit Ans. <br />and Direction of \(\overrightarrow {AB}\) is tan\(\theta\) = \(\frac{y - component}{x - component}\) = \(\frac{1}{-2}\) = \(tan(tan^{-1})\frac{-1}{2}\)<sup>. </sup><br />\(\therefore\) \(\theta\) = \(tan^{-1}\frac{-1}{2}\) Ans.</p>

Q12:

The vector \(\overrightarrow {PQ}\) displaces P(1 , 1) to Q(2 , 5) . Express \(\overrightarrow {PQ}\) in the column vector.  Under what condition the following vectors joining the point A and B \(\overrightarrow {AB}\) are equal to the vector \(\overrightarrow {PQ}\) or -\(\overrightarrow {PQ}\) = \(\overrightarrow {AB}\) ?. Explain.
A (0 , 2) and B (1 , 6)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Here , given point , <br />P9x1 , y1) = (1 , 1) and Q(x<sub>2</sub> , y<sub>2</sub>) = (2 , 5)<br />Now , x - component of \(\overrightarrow {PQ}\) = x<sub>2</sub> - x<sub>1</sub> = 2 - 1 = 1<br />y - component of \(\overrightarrow {PQ}\) = y<sub>2</sub> - y<sub>1</sub> = 5 - 1 = 4<br /><br />\(\therefore\) = \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\) Ans. <br />Direction of \(\overrightarrow {PQ}\) is tan \(\theta\) = \(\frac{4}{1}\) = tan(tan<sup>-1</sup>4) <br />\(\therefore\) \(\theta\) = tan<sup>-1</sup>4</p>

Q13:

The vector \(\overrightarrow {PQ}\) displaces P(1 , 1) to Q(2 , 5) . Express \(\overrightarrow {PQ}\) in the column vector. Under what condition the following vectors joining the point A and B\(\overrightarrow {AB}\) are equal to the vector \(\overrightarrow {PQ}\) or -\(\overrightarrow {PQ}\) = \(\overrightarrow {AB}\) ? . Explain.
A(3 , -3)  , B(4 , 1)


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Given points , <br />A(x<sub>1</sub> , y<sub>1</sub>) = (3 , -3) and B(x<sub>2</sub> , y<sub>2</sub>) = (4 , 1)<br />Now , x - component of \(\overrightarrow {AB}\) = x<sub>2</sub> - x<sub>1</sub> = 4 - 3 = 1<br />y - component of \(\overrightarrow {AB}\) = y<sub>2</sub> - y<sub>1</sub> = 1 - (-3) = 1 + 3 = 4<br />\(\therefore\) \(\overrightarrow {AB}\) = \(\begin {pmatrix} 1 \\ 4 \end{pmatrix}\) but \(\overrightarrow {PQ}\) = \(\begin {pmatrix} 1 \\ 4 \end{pmatrix}\) and tan\(\theta\) = \(\frac{4}{1}\) = tan<sup>-1 </sup>4 = tan(tan<sup>-1</sup>4) \(\theta\) = tan<sup>-1</sup>4<br />\(\therefore\) \(\overrightarrow {AB}\) = \(\overrightarrow {PQ}\) . Ans.</p>

Videos

Magnitude and direction of vector sums | Vectors | Precalculus | Khan Academy
Introduction to Vectors
Magnitude and Direction of a Vector, Example 1
The Socio-economic Activites of Australia

The Socio-economic Activites of Australia

People

The population of Australia is a mixed population. Aborigines migrated about 30,000 years ago from South-east Asia is now in the minority- only about 1%. Their life style is typically distinct. They have preserved their rich culture. The ‘White Australia Policy’ is no more. In the recent past people from non-European nations also came as immigrants. They have highly modernized living standard. Out of total 22.5 million Australian populations, more than 92% live in cities concentrated mainly on the coasts in the south-west and south-east, including the ones in Tasmania. In this sense, Australia is the world’s most urbanized country.

Sheep farm

Australia has very wide semi-arid areas, ideal for live stock farms. Sheep farms are important particularly in Western Australia, New South Wales, and Victoria. Several thousands of sheep are owned by a farmer, who uses motorcycles to graze the herd of sheep and the farm is so big that he uses light air-crafts to fly from one part to another. Cattle are raised in all states and territories, the semi-arid grasslands are ideal for pasturing. These livestock farms are known as ‘stations’. Australia is the largest exporter of wool. World’s 30% wool comes from here. It earns about 6% of the country’s income.

Agriculture

The vast majority of the land is flat, dry and hot. Only about 6% is under crop and fodder cultivation. Still the country has much surplus of agro-products. Wheat is very important. It occupies nearly half of the cultivated area. It uses modern machinery and technology so the production is very high, 70% of which is exported. Paddy is grown in the irrigated area in the south and in the north. Sugarcane is cultivated especially in the coast of Queensland. Oats, barley, maize, oil seeds, tobacco, cotton, apples, grapes, pears, etc are also grown in good quantities.

Mining and industries

Australia is very rich in minerals. Silver, lead, copper, zinc, gold, uranium and coal are found here. Western Australia produces more than the half a number of metallic minerals. North has many good copper mines. Iron is found everywhere, much in the west. Gold is found in very high volume in the west. Australia produces one-third of the world’s uranium, which is found in the Northern Territory and South Australia. Minerals found 40% of the country’s income. Australia has all kinds of industries. Minerals, wheat, wool, meat, dairy products, machinery, etc. are its major exports.

Tourism

Australia has unique plant and animal life. National Parks like the Uluru and Kakadu attract a large number of tourists. Great Barrier Reef, Sydney, Brisbane beaches, vast deserts and historic sites all allure tourists from throughout the world. The way of life of the aborigines in the interior of the country is also the matter of interest.

Lesson

Our Earth

Subject

Social Studies

Grade

Grade 9

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