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# Tutor profile: Siva T.

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Siva T.
Bachelors in Mechanical Engineering, Masters in Machine Design
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## Questions

### Subject:Geometry

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Question:

Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9) and D(5, 4). What is the shape of the quadrilateral?

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Siva T.

The lengths of the four sides, AB, BC, CD and DA are all equal to 41−−√41. Hence, the given quadrilateral is either a Rhombus or a Square. How to determine whether the quadrilateral is a square or a rhombus? The diagonals of a square are equal. The diagonals of a rhombus are unequal. Compute the lengths of the two diagonals AC and BD. The length of AC is 162−−−√162 and the length of BD is 2–√2. As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus.

### Subject:Algebra

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Question:

Solve (x + 3)(x − 3).

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Siva T.

=x(x − 3) + 3(x − 3) = x2 − 3x + 3x − 9 = x2 − 9

### Subject:Physics

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Question:

A golfer practising on a range with an elevated tee 4.9 m above the fairway is able to strike a ball so that it leaves the club with a horizontal velocity of 20 m s–1. (Assume the acceleration due to gravity is 9.80 m s–2, and the effects of air resistance may be ignored unless otherwise stated.) a How long after the ball leaves the club will it land on the fairway? b What horizontal distance will the ball travel before striking the fairway? c What is the acceleration of the ball 0.5 s after being hit? d Calculate the speed of the ball 0.80 s after it leaves the club. e With what speed will the ball strike the ground?

Inactive
Siva T.

x = ut + 0.5at2 then 4.9 m = 0 + 0.5(9.8 m s–2)t 2 and t = 1.0 s b x = (average speed)(time) = (20 m s–1)(1.0 s) = 20 m c The acceleration of the ball is constant at any time during its flight, and is equal to the acceleration due to gravity = 9.8 m s–2 down d After 0.80 s, the ball has two components of velocity: vx = 20 m s–1 and vy = 0 + (9.8 m s–2)(0.80 s) = 7.84 m s–1 The speed of the ball at 0.80 s is given by: [(20 m s–1) 2 + (7.84 m s–1) 2 ] ½ = 21.5 m s–1 The ball will strike the ground 1.0 s after it is struck. Then vx = 20 m s–1 and vy = 0 + (9.8 m s–2)(1.0 s) = 9.8 m s–1 The speed of the ball at 1.0 s is given by: [(20 m s–1) 2 + (9.8 m s–1) 2 ] ½ = 22.3 m s–1

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