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Oksanka [162]
2 years ago
12

The hammer throw was one of the earliest Olympic events. In this event, a heavy ball attached to a chain is swung several times

in a circular path until it is released. The winning athlete is the one who throws the ball the greatest distance. The last complete rotation of 2016 Olympic champion Anita Włodarczyk’s final turn took only 0.43 s. The radius of the ball’s path, including her extended arms, was 2.1 m. a) What was the frequency of this rotation? b) What was the speed of the ball? c) What was the ball’s acceleration, in units of g?
Physics
1 answer:
Aleonysh [2.5K]2 years ago
7 0

Answer:

Given that

T= 0.43 s

Radius of the ball path's , r=2.1 m

a)

We know that

f= 1/T

Here f= frequency

      T= Time period

Now by putting the values

f= 1/T

T= 0.43 s

f= 1/0.43

f=2.32 Hz

b)

We know that

V= ω r

ω = 2 π f

ω=Angular speed

V= Linear speed

ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s

V= ω r= 14.60 x 2.1 = 30.66 m/s

c)

Acceleration ,a

a =ω ² r

a= 14.6 ² x 2.1 = 447.63 m/s²

We know that g = 10 m/s²

So

a= a/g= 447.63/10 = 44.7 g m/s²

a= 44.7 g m/s²

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b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

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b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

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\Delta x = x(4)-x(0)=20.8-0=20.8 m

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So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

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While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

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