Chapter 9, Problem 41P

(a)

To determine

The time taken by the water to travel from the nozzle to the ground.

Answer to Problem 41P

The time taken by the water to travel from the nozzle to the ground is $0.533\text{s}$.

Explanation of Solution

Formula to calculate the time taken by the water to reach the ground is,

$t=\sqrt{\frac{2\left(\Delta y\right)}{a_y}}$

• $t$ is taken time,
• $\left(\Delta y\right)$ is the distance
• $a_y$ is the acceleration along upward direction.

Substitute $\left(-1.5\text{m}\right)$ for $\left(\Delta y\right)$, $\left(-9.8{\text{m/s}}^{\text{2}}\right)$ for $a_y$ to find $t$

$\begin{array}{c}t=\sqrt{\frac{2\left(-1.5\text{m}\right)}{\left(-9.8{\text{m/s}}^{\text{2}}\right)}}\\ =0.533\text{s}\end{array}$

Thus, the time taken by the water to travel from the nozzle to the ground is $0.533\text{s}$.

Conclusion:

Therefore, time taken by the water to travel from the nozzle to the ground is $0.533\text{s}$.

(b)

To determine

The speed that must have the stream while leaving the nozzle.

Answer to Problem 41P

The speed that must have the stream while leaving the nozzle is $14.5\text{m/s}$.

Explanation of Solution

Formula to calculate the speed of the stream is,

$v_{nozzle}=\frac{\left(\Delta x\right)}{t}$

• $t$ is taken time,
• $\left(\Delta x\right)$ is the horizontal range
• $v_{nozzle}$ is the speed of the stream at the nozzle

Substitute $\left(8\text{m}\right)$ for $\left(\Delta x\right)$, $0.533\text{s}$ for $t$ to find $v_{nozzle}$

$\begin{array}{c}{v}_{nozzle}=\frac{\left(8\text{}\text{m}\right)}{\left(0.533\text{s}\right)}\\ =14.5\text{m/s}\end{array}$

Thus, the speed that must have the stream while leaving the nozzle is $14.5\text{m/s}$.

Conclusion:

Therefore, the speed that must have the stream while leaving the nozzle is $14.5\text{m/s}$.

(c)

To determine

The speed at which the plunger must be moved by using continuity equation.

Answer to Problem 41P

The speed at which the plunger must be moved by using continuity equation is $0.145\text{m/s}$.

Explanation of Solution

Formula to calculate the speed of the plunger from continuity equation is,

$v=\left(\frac{\pi r_s^2}{\pi r_l^2}\right)v_{nozzle}$

• $v$ is the speed of the plunger.,
• $r_s\text{and}{r}_l$ are the radius of the smaller and the larger tube.
• $v_{nozzle}$ is the speed of the stream at the nozzle

From unit conversion,

$1\text{cm}=10\text{mm}$

Substitute $\left(\text{1mm}\right)$ for $r_s$, $1\text{cm}$ for $r_l$, $14.5\text{m/s}$ for $v_{nozzle}$ to find $v$

$\begin{array}{c}v=\frac{\left(\pi {\left(1\text{mm}\right)}^2\right)}{\left(\pi \left(1\text{cm}\right)\left(\frac{1\times 10\text{mm}}{1\text{cm}}\right)\right)}\left(14.5\text{m/s}\right)\\ =0.145\text{m/s}\end{array}$

Thus, the speed at which the plunger must be moved by using continuity equation is $0.145\text{m/s}$.

Conclusion:

Therefore, the speed at which the plunger must be moved by using continuity equation is $0.145\text{m/s}$.

(d)

To determine

The pressure of the nozzle.

Answer to Problem 41P

The pressure of the nozzle is $\text{1}\text{.013}\times {\text{10}}^5\text{Pa}$.

Explanation of Solution

The pressure at the nozzle is the atmospheric pressure.

So, the pressure at the nozzle is

$\begin{array}{c}{P}_{nozzle}=P_{atm}\\ =1.013\times {10}^5\text{Pa}\end{array}$

Thus, the pressure of the nozzle is $\text{1}\text{.013}\times {\text{10}}^5\text{Pa}$.

Conclusion:

Therefore, the pressure of the nozzle is $\text{1}\text{.013}\times {\text{10}}^5\text{Pa}$.

(e)

To determine

The pressure needed in the larger cylinder by using Bernoulli’s equation.

Answer to Problem 41P

The pressure needed in the larger cylinder is $2.06\times {\text{10}}^5\text{Pa}$.

Explanation of Solution

Formula to calculate the pressure needed in the larger cylinder is,

$P=P_{atm}+\frac{{\rho }_{water}}{2}\left(v_{nozzle}^2-v^2\right)$

• $P$ is the pressure in the larger cylinder
• $P_{atm}$ is the atmospheric pressure
• ${\rho }_{water}$ is the density of water

Substitute, $1.013\times {10}^5\text{Pa}$ for $P_{atm}$, $1\times {10}^3{\text{kg/m}}^{\text{3}}$ for ${\rho }_{water}$, $14.5\text{m/s}$ for $v_{nozzle}$, $0.145\text{m/s}$ for $v$ to find $P$,

$\begin{array}{c}P=\left(1.013\times {10}^5\text{Pa}\right)+\frac{\left(1\times {10}^3{\text{kg/m}}^{\text{3}}\right)}{2}\left({\left(14.5\text{m/s}\right)}^2-{\left(0.145\right)}^2\right)\\ =2.06\times {10}^5\text{Pa}\end{array}$

Thus, the pressure needed in the larger cylinder is $2.06\times {\text{10}}^5\text{Pa}$.

Conclusion:

Therefore, the pressure needed in the larger cylinder is $2.06\times {\text{10}}^5\text{Pa}$.

(f)

To determine

The force that must be exerted on the trigger to achieve the desired range.

Answer to Problem 41P

The force that must be exerted on the trigger to achieve the desired range is $33\text{N}$.

Explanation of Solution

Formula to calculate the pressure needed in the larger cylinder is,

$\begin{array}{c}F=\left(\Delta P\right)\left(\pi r_l^2\right)\\ =\left(P-P_{atm}\right)\left(\pi r_l^2\right)\end{array}$

• $P$ is the pressure in the larger cylinder
• $P_{atm}$ is the atmospheric pressure
• $r_l$ is the radius of the larger tube.

From unit conversion is,

$1\text{cm}=1\times {10}^{-2}\text{m}$

Substitute, $1.013\times {10}^5\text{Pa}$ for $P_{atm}$, $\text{2}\text{.06}\times {\text{10}}^5\text{Pa}$ for $P$, $1\text{cm}$ for $r_l$ to find $F$,

$\begin{array}{c}F=\left[\left(2.06\times {10}^5\text{Pa}\right)-\left(1.013\times {10}^5\text{Pa}\right)\right]\left(\pi {\left\{\left(1\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}}^2\right)\\ =33\text{N}\end{array}$

Thus, the force that must be exerted on the trigger to achieve the desired range is $33\text{N}$.

Conclusion:

Therefore, the force that must be exerted on the trigger to achieve the desired range is $33\text{N}$.