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3 years ago

6

when a stationary rugby ball is kicked, it is contact with a player's about for 0.05 s. during this short time, the ball acceler

ates at 600m/s/s.calculate the speed at which the ball leaves the player's boot

1 answer:

Ad libitum [116K]3 years ago

3 0

Answer:

30 m/s

Explanation:

Applying,

v = u+at................ Equation 1

Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.

From the question,

Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s

Substitute these values into equation 5

v = 0+(600×0.05)

v = 30 m/s

Hence the speed at which the ball leaves the player's boot is 30 m/s

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Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

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$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

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$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

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