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3 years ago

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What is the force of a lead ball that starts with a momentum of 7 kgm/s and end with a momentum of 22 kgm/s over a time of 31

seconds?
Explain how you got your answer.

1 answer:

Gwar [14]3 years ago

6 0

Answer:

F = 0.483 N

Explanation:

Initial momentum, $P_i= 7\ kg-m/s$

Final momentum, $P_f= 22\ kg-m/s$

Time, t = 31 s

We need to find the force of a lead ball. We can use here the impulse momentum theorem.

$\text{Change in momentum}=F\times t$

F is force

$P_f-P_i=Ft\\\\F=\dfrac{P_f-P_i}{t}\\\\F=\dfrac{22-7}{31}\\\\F=0.483\ N$

So, the force is 0.483 N.

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A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

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$h_1 = 0$ is the heigth of the lower section

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$g=9.8 m/s^2$ is the acceleration of gravity

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$v_2$ is hte speed in the higher section

We can re-write the equation as

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Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

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Now we can use equation (2) to find the speed in the lower section:

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Substituting

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Therefore, the volume flow rate is

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