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Natasha_Volkova [10]

3 years ago

14

If you dribble a basketball with a frequency of 1.77 Hz, how long does it take for you to complete 12 dribbles?

1 answer:

Licemer1 [7]3 years ago

8 0

<h2>It takes 6.78 seconds to complete 12 dribbles.</h2>

Explanation:

Frequency of dribble = 1.77 Hz

That is

Number of dribbles in 1 second = 1.77

$\texttt{Time taken for 1 dribble = }\frac{1}{1.77}=0.565s$

Now we need to find how long does it take for you to complete 12 dribbles.

Time taken for 12 dribbles = 12 x Time taken for 1 dribble

Time taken for 12 dribbles = 12 x 0.565

Time taken for 12 dribbles = 6.78 seconds

It takes 6.78 seconds to complete 12 dribbles.

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A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

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Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

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$= 44.23 m/s$

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$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

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$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

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At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

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