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3 years ago

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A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. How far did t

he ball travel before it hit the ground? Round the answer to the nearest meter.

2 answers:

Furkat [3]3 years ago

3 0

What are your answers?

Agata [3.3K]3 years ago

3 0

Answer:

60m (to nearest meter)

Explanation: How far the ball traveled refers to the range covered by the ball before it hits the ground. For an object projected at an angle of ∝ to the horizontal, the range, R covered is given by;

$R=u^{2}sin2\alpha/2g$

Where u is the initial velocity and g is acceleration due to gravity.

$u=26mls\\\alpha=30^{o}\\g=9.81m/s^{2}$

The value of g could be safely assumed to be $9.8m/s^{2}$

Therefore;

$R=26^{2}sin2(30)/9.8$

$R=(26*26*sin60)/9.8$

$R=676*0.8660/9.8$

$R=59.68m$ which is approximately 60m to nearest meter.

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A car driving at 20 m/s accelerates continuously 2m/s2 for 3 seconds. What is its final velocity

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Answer: V= u+ at

V= final velocity

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a=acceleration

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Explanation:

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2 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

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3 years ago

Please tell me the answer

maria [59]

Answer:

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2 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

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$\tau = I \times \alpha$

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So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

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Hence, $\tau_{1} = \tau_{2}$

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$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

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3 years ago

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