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levacccp [35]
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​
Physics
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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3 years ago
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

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.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

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

Where

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

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

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

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

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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