Question

Maria is filling a bucket of water from a faucet. After she turns it on, she sees that the cross-sectional area of the water stream right below the faucet is about the size of a quarter, or 4.62×10^4 m2. The stream constricts to about the size of a dime, or 2.52×10^4 m2, after falling 2.50 cm. If the bucket Maria is filling has a volume of 10.0 L, how long will it take to fill the bucket?

Answer 1

Answer:

The time is $4.76\times10^{-7}\ sec$

Explanation:

Given that,

Area $A_{1}=4.62\times10^{4}\ m^2$

Area $A_{2}=2.52\times10^{4}\ m^2$

Height = 2.50 cm

Volume = 10.0 L

We need to calculate the speed

Using equation of continuity

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

Put the value into the formula

$4.62\times10^{4}\times v_{1}=2.52\times10^{4}\times v_{1}$

$4.62v_{1}=2.52v_{2}$.....(I)

$v_{1}=\dfrac{2.52}{4.62}v_{2}$

$v_{1}=0.545v_{2}$

Now, using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rhi\times v_{1}^2+\rho gh=P_{2}+\dfrac{1}{2}\rhi\times v_{2}^2$

Here, $P_{1}=P_{2}=P_{atm}$

$v_{2}^2=v_{1}^2+2gh$.....(II)

Put the value $v_{1}$ into the formula

$v_{2}^2=(0.545v_{2})^2+2\times9.8\times2.50\times10^{-2}$

$v_{2}^2=0.297v_{2}^2+0.49$

$v_{2}^2(1-0.297)=0.49$

$v_{2}=\sqrt{\dfrac{0.49}{0.703}}$

$v_{2}=0.835\ m/s$

Put the value of $v_{2}$ in the equation (I)

$v_{1}=0.545\times0.835$

$v_{1}=0.46\ m/s$

We need to calculate the flow rate

Using formula of flow rate

$Q=A_{1}v_{1}$

$Q=(4.62\times10^{4})\times0.46$

$Q=2.1\times10^{4}\ m^3/s$

We need to calculate the time

Using formula of time

$t = \dfrac{V}{Q}$

Put the value into the formula

$t=\dfrac{10.0\times10^{-3}}{2.1\times10^{4}}$

$t=4.76\times10^{-7}\ sec$

Hence, The time is $4.76\times10^{-7}\ sec$

Answer 2

Answer:

t = 47.62 sec

Explanation:

Given data;

$A_1 = 4.62 \times 10^4 m^2$

$A_2 = 2.52 \times 10^4 m^2$

h = 2.50 cm

volume 10 L

from

$A_1 v_1 = A_2 v_2$

$4.62 \times 10^4 v_1 = 2.52 \times 10^4 v_2$

$4.62 v_1 = 2.52 v_2$ ......1

from bernoulli eq

$P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2$

$P_1 =P_2 = P_{atm}$

$v_2^2 = v_1^2 +2gh$ ... 2

from 1 and 2 equation

$v_1 = 0.46 m/s$

volume flow rate is

$Q = A_1 \times v_1 = 4.62 \times 10^[-4} v_1 = 2.1 \times 10^{-4} m^3/s$

$t = \frac{v}{Q}$

$t =\frac{10\times 10^{-3}}{2.1 \times 10^{-4}} = 47.62 s$