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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 439: Lesson Check

Exercise 39

Step 1

1 of 2

When we partially block the end of the garden hose with our thumb we reduce the effective area the fluid can flow through.

As dictated by the Equation of Continuity this brings about an increase in the speed of the fluid, since

$$
begin{align*}
v_1 cdot A_1 = v_2 cdot A_2
end{align*}
$$

So we conclude that the speed of the water coming out of the nozzle is greater than the speed in the hose.

Result

2 of 2

The speed of the water coming out of the nozzle is greater than the speed in the hose.

Exercise 40

Step 1

1 of 2

When we blow air across the top of the paper we reduce the pressure there according to the Bernoulli’s principle, since the air moves with higher speed.

This reduction in pressure leads to a net upward force that lifts the paper.

Result

2 of 2

By blowing air we reduce the pressure there, this results in a net upward force.

Exercise 41

Step 1

1 of 2

A real fluid flowing past a stationary surface experiences frictional forces. The tendency of fluids to resist flow is referred to as viscosity.

Just as work against friction must be done to move a block on a rough surface, so work must be done to push a real fluid through a pipe.

Result

2 of 2

Work is required because of friction between the fluid and the surface of the tube.

Exercise 42

Solution 1

Solution 2

Step 1

1 of 2

As the fluid falls from the faucet its speed increases due to the force of gravity.

Since the Equation of Continuity states that:

$$
begin{equation*}
v cdot A = text{const}
end{equation*}
$$

We see that an increase in the speed of the fluid will bring about a decrease in the effective area of the fluid flow. Hence the stream becomes narrower as it falls.

Result

2 of 2

As the fluid falls its speed increases, so the area of the stream must decrease owing to the Equation of Continuity.

Step 1

1 of 2

A stream of water emerging from a water faucet becomes narrower as it falls because its speed keeps on increasing with it height of fall due to acceleration due to gravity.
Using the equation of continuity for better understanding:

$$
A_1v_1 = A_2v_2
$$

Where the initial cross sectional area of the flow of water is $A_1$,

Initial speed of water is $v_1$,

Final cross sectional area of the flow of water is $A_2$,

Final speed of water is $v_2$,

Therefore, if velocity of the flow increases then area gets reduced and thus the stream of water becomes narrower.

Result

2 of 2

Therefore, if velocity of the flow increases then area gets reduced and thus the stream of water becomes narrower.

Exercise 43

Solution 1

Solution 2

Step 1

1 of 2

When the diameter of the pipe through which the fluid flows is increased by a factor of 3, the area of that pipe is increased by a factor of 9.

Since the Equation of Continuity states that:

$$
begin{equation*}
v cdot A = text{const}
end{equation*}
$$

We see that an increase in the area of the flow of the fluid by a factor of 9 will bring about a decrease in the speed by a factor of 9.

Hence the final speed is $frac{v}{9}$

Result

2 of 2

The final speed is $frac{v}{9}$, owing to the Equation of Continuity.

Step 1

1 of 2

Using the equation of continuity for the pipe:

$$
A_1v_1 = A_2v_2
$$

Where the cross sectional area of the hose is $A_1 = pi dfrac{d^2}{4}$,

Speed of water is $v_1 = v$,

New area of the pipe is $A_2 = pi dfrac{(3d)^2}{4}$,

Therefore,

$$
begin{align*}
A_1v_1 &= A_2v_2\
\
v_2 &= dfrac{A_1v_1}{A_2}\
\
v_2 &= dfrac{pi dfrac{d^2}{4} cdot v}{pi dfrac{(3d)^2}{4}}\
\
v_2 &= dfrac{d^2 cdot v}{9d^2}\
\
v_2 &= boxed{dfrac{v}{9}}\
end{align*}
$$

Therefore the new speed of the fluid is equal to $dfrac{v}{9}$, because if the diameter of the pipe is increased 3 times, then its area would increase nine time and according to the equation of continuity, its velocity would decrease nine times.

Result

2 of 2

Therefore the new speed of the fluid is equal to $dfrac{v}{9}$

Exercise 44

Step 1

1 of 3

### Knowns

– The first cross-sectional area of the hose $A_1 = 0.0075text{ m}^2$

– The speed of water through the hose $v_1 = 1.3 ; frac{text{m}}{text{s}}$

– The second cross-sectional area of the hose $A_2 = 0.0033text{ m}^2$

Step 2

2 of 3

### Calculation

Using the Equation of Continuity we can find the new speed $v_2$:

$$
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
$$

Rearranging for $v_2$ we have:

$$
begin{align*}
v_2 = v_1 cdot frac{A_1}{A_2}
end{align*}
$$

Plugging in the values we have:

$$
begin{align*}
v_2 &= 1.3 ; frac{text{m}}{text{s}} cdot frac{0.0075text{ m}^2}{0.0033text{ m}^2} \
v_2 &= 2.95 ; frac{text{m}}{text{s}}
end{align*}
$$

Result

3 of 3

The new speed of the water is $v_2 = 2.95 ; frac{text{m}}{text{s}}$

Exercise 45

Solution 1

Solution 2

Step 1

1 of 4

Using the continuity equation, we can determine the new cross sectional area

$$
A_{1}v_{1} = A_{2}v_{2}
$$

Step 2

2 of 4

a. Based on the equation, both sides should be equal, hence, if one variable in one side, increases, the other variable should decrease and vise versa.

In the problem, the speed of the water decreased, therefore, the cross-sectional area is expected to increase.

Step 3

3 of 4

b. Rearranging the equation to get the new cross-sectional area,

$$
A_{2} = dfrac{A_{1}V_{1}}{V_{2}} = dfrac{(0.0084m^2)(2.5m/s)}{1.1m/s}
$$

$$
A_{2} = 0.019m^2
$$

Result

4 of 4

a. The new cross-sectional area is greater than $0.0084m^2$

b. $A_{2} = 0.019m^2$

Step 1

1 of 3

### Knowns

– The first cross-sectional area of the hose $A_1 = 0.0084 text{ m}^2$

– The speed of water through the hose $v_1 = 2.5 ; frac{text{m}}{text{s}}$

– The new speed of the water $v_2 = 1.1 ; frac{text{m}}{text{s}}$

Step 2

2 of 3

section*{Calculation}
begin{enumerate}[a)]
item
We know from the Continuity Equation that $v cdot A = text{const}$ \
Since our speed decreases, the area must increase, so $A_2 > 0.0084text{ m}^2$
item
Using the Equation of Continuity we can find the required cross-sectional area of the hose $A_2$:
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
Rearranging for $A_2$ we have:
begin{align*}
A_2 = A_1 cdot frac{v_1}{v_2}
end{align*}
Plugging in the values we have:
begin{align*}
A_2 &= 0.0084text{ m}^2 cdot frac{2.5 ; frac{text{m}}{text{s}}}{1.1 ; frac{text{m}}{text{s}}} \
A_2 &= 0.0191text{ m}^2
end{align*}
end{enumerate}

Result

3 of 3

begin{enumerate}[a)]
item
The new cross-sectional area of the hose must increase owing to the Equation of Continuity, so $A_2$ is greater than $0.0084text{ m}^2$
item
$A_2 = 0.0191text{ m}^2$
end{enumerate}

Exercise 46

Step 1

1 of 3

### Knowns

– The diameter of the pipe $d_1 = 27text{ cm}$

– The speed of the fluid through the pipe $v_1 = 1.9 ; frac{text{m}}{text{s}}$

– The final speed of the fluid $v_2 = 3.1 ; frac{text{m}}{text{s}}$

Step 2

2 of 3

### Calculation

Using the Equation of Continuity we can find the required diameter $d_2$:

$$
begin{equation*}
v_1 , A_1 = v_2 , A_2
end{equation*}
$$

Where the areas are found as follows:

$$
begin{align*}
A_1 &= left(frac{d_1}{2} right)^2 , pi \
A_2 &= left(frac{d_2}{2} right)^2 , pi
end{align*}
$$

So we have:

$$
begin{align*}
v_1 , d_1^2 = v_2 , d_2^2
end{align*}
$$

Rearranging for $d_2$ we have:

$$
begin{align*}
d_2 = d_1 cdot sqrt{frac{v_1}{v_2}}
end{align*}
$$

Plugging in the values we have:

$$
begin{align*}
d_2 &= 2.7text{ cm} cdot sqrt{frac{1.8 ; frac{text{m}}{text{s}}}{3.1 ; frac{text{m}}{text{s}}}} \
d_2 &approx 2.06text{ cm}
end{align*}
$$

Result

3 of 3

The new diameter of the pipe is $d_2 = 2.06text{ cm}$

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