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3 years ago

Uber drivers are considered 'independent contractors' and thus can apply for business licenses.

2 answers:

8 0

True, Uber drivers can be deactivated/fired if they ratings reach a certain point, despite the fact they are still considered “independent contractors”. They still belong to a company.

Natasha2012 [34]3 years ago

4 0

I’m pretty sure it’s true!

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A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well

ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

= 14.9 * 20.1 = 299.5 m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T* t )

= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

= 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = $\frac{Q}{4\pi T} * W (u)$

where : Q = 2725.5

T = 299.5

W(u) = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0

3 years ago

Problem 1: A catchment has the following Horton’s infiltration parameters: f0=280 mm/hr, fc=25 mm/hr and k = 2.5 hr-1 . For the

harkovskaia [24]

Answer:

Ponding will occur in 40mins

Explanation:

We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.

Consider checking attachment for the step by step solution.

6 0

3 years ago

Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro

sasho [114]

Answer:

Just answered this to confirm my profile.

Explanation:

I dont have a clue, this is just to confirm my profile.

8 0

3 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0

3 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago