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

The time is
`4.76*10^(-7)\ sec`

Step-by-step explanation:

Given that,

Area
`A_(1)=4.62*10^(4)\ m^2`

Area
`A_(2)=2.52*10^(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*10^(4)* v_(1)=2.52*10^(4)* v_(1)`

`4.62v_(1)=2.52v_(2)`.....(I)

`v_(1)=(2.52)/(4.62)v_(2)`

`v_(1)=0.545v_(2)`

Now, using Bernoulli equation

`P_(1)+(1)/(2)\rhi* v_(1)^2+\rho gh=P_(2)+(1)/(2)\rhi* 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*9.8*2.50*10^(-2)`

`v_(2)^2=0.297v_(2)^2+0.49`

`v_(2)^2(1-0.297)=0.49`

`v_(2)=\sqrt{(0.49)/(0.703)}`

`v_(2)=0.835\ m/s`

Put the value of
`v_(2)` in the equation (I)

`v_(1)=0.545*0.835`

`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*10^(4))*0.46`

`Q=2.1*10^(4)\ m^3/s`

We need to calculate the time

Using formula of time

`t = (V)/(Q)`

Put the value into the formula

`t=(10.0*10^(-3))/(2.1*10^(4))`

`t=4.76*10^(-7)\ sec`

Hence, The time is
`4.76*10^(-7)\ sec`

Answer 2

t = 47.62 sec

Step-by-step explanation:

Given data;

`A_1 = 4.62 * 10^4 m^2`

`A_2 = 2.52 * 10^4 m^2`

h = 2.50 cm

volume 10 L

from

`A_1 v_1 = A_2 v_2`

`4.62 * 10^4 v_1 = 2.52 * 10^4 v_2`

`4.62 v_1 = 2.52 v_2` ......1

from bernoulli eq

`P_1 + (1)/(2) \rho v_1^2 + \rho g h = P_2 + (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 * v_1 = 4.62 * 10^[-4} v_1 = 2.1 * 10^(-4) m^3/s`

`t  = (v)/(Q)`

`t =(10* 10^(-3))/(2.1 * 10^(-4)) = 47.62 s`