UNIT HACK FREE PREVIEW

Before you reach for a formula, read the units.

Use target units to guide Chem/Phys calculations, then check the setup against the relevant science.

This free, hands-on sample lets you try four Chem/Phys questions, get feedback as you answer, and see how each solution is built and checked. It gives you a practical look at the approach developed further in the full Unit Hack course.

Units first. Math last.

TargetDecodeCollectBuildCancelCheck

Practice Set: Fluid Delivery System

Choose an answer to see immediate feedback. Open the worked solution to follow each unit transformation at your own pace.

Researchers are evaluating a pump used to deliver saline through a narrow circular tube. During one trial, the pump maintains a constant volume flow rate of 0.80 mL/s for 2.5 minutes. A pressure sensor records a constant 12 kPa pressure difference between the pump outlet and the end of the tube. The internal tube radius at the measurement point is 1.0 mm. The measured saline density is approximately 1.0 g/mL.

Assume the saline is incompressible, the flow is steady, and mechanical energy losses are negligible. The pressure difference represents the pressure against which the pump delivers the measured volume.

01

How much saline does the pump deliver during the trial?

Not quite. The correct answer is C: 120 mL. The 2.5-minute trial lasts 150 seconds. At 0.80 mL per second, the pump delivers 120 mL.

Not quite. The correct answer is C: 120 mL. The 2.5-minute trial lasts 150 seconds. At 0.80 mL per second, the pump delivers 120 mL.

Correct. The 2.5-minute trial lasts 150 seconds. At 0.80 mL per second, the pump delivers 120 mL.

Not quite. The correct answer is C: 120 mL. The 2.5-minute trial lasts 150 seconds. At 0.80 mL per second, the pump delivers 120 mL.

Answer: C. 120 mL

Worked solution

Convert the operating time to seconds, then multiply by the volume flow rate.

Follow the unit setup

Use Next or Previous to review each step.

Target1 / 5
mL

The question asks for total delivered volume.

‹
Decode2 / 5
mLs × s = mL

A flow rate multiplied by elapsed time gives a volume.

Convert3 / 5
2.5 min × 60 s1 min = 150 s

Convert minutes to seconds using a vertical conversion factor.

Cancel4 / 5
0.80 mLs × 150 s1

Seconds cancel. Milliliters remain.

Check5 / 5
0.80 × 150 = 120 mL

150 seconds at 0.80 mL per second gives 120 mL.

›
0.80 mL1 s × 2.5 min1 × 60 s1 min = 120 mL

Math shortcut: 2.5 × 60 = 150; 0.80 × 150 = 8 × 15 = 120.

Unit Hack method

Target
Total volume in mL.
Decode
A rate in mL/s becomes a volume in mL when multiplied by a duration in seconds.
mLs × s = mL
Collect
0.80 mL/s and 2.5 min.
Build
Arrange volume per second with elapsed time and a minutes-to-seconds conversion. The result must end in mL.
Cancel
Minutes and seconds cancel against matching factors in the stacked calculation. Only mL remains.
Check
The result is a volume, not a rate, and the magnitude is reasonable: 0.8 mL each second for 150 seconds is 120 mL.

Proof It Works

The unit-guided setup and the constant-flow definition give the same delivered volume.

Unit Hack setup
mL = (mL/s) × s0.80 mL/s × 150 s
120 mL
Conventional equation
V = Q × t= (0.80 mL/s)(150 s)
120 mL

This check requires the flow rate to remain constant during the trial.

02

Approximately how much mechanical work does the pump do to deliver the volume against the measured pressure difference?

Not quite. The correct answer is B: 1.44 J. The delivered volume is 120 mL. Under the stated assumptions, pressure difference multiplied by delivered volume gives 1.44 J.

Correct. The delivered volume is 120 mL. Under the stated assumptions, pressure difference multiplied by delivered volume gives 1.44 J.

Not quite. The correct answer is B: 1.44 J. The delivered volume is 120 mL. Under the stated assumptions, pressure difference multiplied by delivered volume gives 1.44 J.

Not quite. The correct answer is B: 1.44 J. The delivered volume is 120 mL. Under the stated assumptions, pressure difference multiplied by delivered volume gives 1.44 J.

Answer: B. 1.44 J

Worked solution

First determine the delivered volume from the constant flow rate and duration: 0.80 × 150 = 120 mL. Convert it to cubic meters and convert the pressure difference to pascals.

Follow the unit setup

Use Next or Previous to review each step.

Target1 / 5
J = N · m

The target quantity is mechanical work.

‹
Decode2 / 5
Nm2 × m3

Pressure times volume has the dimensions of work.

Collect3 / 5
12 kPa→1.2 × 104 Pa
120 mL→1.2 × 10−4 m3

Convert the pressure difference and delivered volume.

Cancel4 / 5
Nm2 × m2 · m = N · m

Expand cubic meters as square meters times meters and cancel matching units.

Check5 / 5
(1.2 × 1.2) × 104 + (−4)=1.44 J

The assumed pressure-work relationship supports this setup.

›
120 mL × 10−6 m31 mL = 1.2 × 10−4 m3
12 kPa × 103 Pa1 kPa = 1.2 × 104 Pa
(1.2 × 104 Pa) × (1.2 × 10−4 m3) = 1.44 J

Math shortcut: Multiply 1.2 × 1.2 and add the exponents: 104 × 10−4 = 1.

Unit Hack method

Target
Work in joules (J).
Decode
A joule is a newton-meter; a pascal is a newton per square meter. Pressure multiplied by volume therefore has the units of work.
J = N · m = Nm2 × m3 = Pa · m3
Matching units provides a possible setup, not proof of the physical relationship.
Collect
Pressure difference: 12 kPa. Flow rate: 0.80 mL/s. Time: 2.5 min. The flow rate and duration give 120 mL delivered.
0.80 mL/s × 2.5 min × 60 s/min = 120 mL
Convert pressure to pascals and volume to cubic meters:
12 kPa = 1.2 × 104 Pa
120 mL = 1.2 × 10−4 m3
Build
The target can be written as Pa · m3. Multiply the pressure difference by the delivered volume to match those units.
(1.2 × 104 Pa) × (1.2 × 10−4 m3)
The passage specifies a constant opposing pressure difference acting over the delivered volume; under those conditions, the product represents work.
Cancel
First, the conversion factors cancel kPa and mL. Expand Pa as N/m2 and m3 as m2 · m. Matching square-meter units cancel, leaving N · m, or J.
Nm2 × m2 · m = N · m = J
Check
Multiply the coefficients and combine the powers of ten.
(1.2 × 1.2) × 104 + (−4) = 1.44 J
The scale is reasonable (about 1 J), the units are joules, and the passage assumptions support pressure work. Answer: B.

Proof It Works

The unit-guided construction can be checked using force, displacement, and pressure.

Unit Hack setup
J = Pa · m³(12 × 10³ Pa)(1.2 × 10⁻⁴ m³)
1.44 J
Conventional equation
W = Fd = (ΔP · A)d= ΔP · V = (12 × 10³)(1.2 × 10⁻⁴) J
1.44 J

Here V = Ad for an equivalent fluid displacement. The constant opposing pressure difference and negligible-loss assumptions make the pressure-work relationship applicable.

03

What is the approximate mean speed of saline through the tube at the measurement point?

Not quite. The correct answer is B: approximately 0.25 m/s. Flow rate divided by circular cross-sectional area gives mean speed. The area comes from A = πr².

Correct. Flow rate divided by circular cross-sectional area gives mean speed. The area comes from A = πr².

Not quite. The correct answer is B: approximately 0.25 m/s. Flow rate divided by circular cross-sectional area gives mean speed. The area comes from A = πr².

Not quite. The correct answer is B: approximately 0.25 m/s. Flow rate divided by circular cross-sectional area gives mean speed. The area comes from A = πr².

Answer: B. Approximately 0.25 m/s

Worked solution

First use the circle-area relationship to obtain the cross-sectional area, then use the units to organize the flow rate per area into speed.

Follow the unit setup

Use Next or Previous to review each step.

Target1 / 5
ms

Mean fluid speed is a distance traveled per unit time.

‹
Decode2 / 5
m3s · m2 = ms

Volume flow rate divided by cross-sectional area has speed units.

Collect3 / 5
r = 1.0 × 10−3 m

Convert millimeters to meters. Geometry supplies the area relationship.

Build4 / 5
A = πr2 = 3.14 × 10−6 m2

Use the circular cross-sectional area, not radius alone.

Cancel5 / 5
0.80 × 10−6 m2 · m3.14 × 10−6 m2 · s ≈ 0.25 m/s

Cancel matching square meters; the speed is approximately 0.25 m/s.

›
0.80 mL/s × 10−6 m31 mL = 0.80 × 10−6 m3/s
1.0 mm × 10−3 m1 mm = 1.0 × 10−3 m
A = πr2 = 3.14 × (1.0 × 10−3 m)2 = 3.14 × 10−6 m2
v = 0.80 × 10−6 m2 · m/s3.14 × 10−6 m2 ≈ 0.25 m/s

Math shortcut: The 10−6 factors divide out. Approximate 3.14 as 3.2; 0.8 ÷ 3.2 = 0.25.

Unit Hack method

Target
Mean fluid speed in m/s.
Decode
Volume flow rate has units m³/s; cross-sectional area has units m². Dividing gives m/s.
Collect
Flow rate: 0.80 mL/s; tube radius: 1.0 mm.
Build
Dividing m³/s by m² yields m/s. But the tube radius is a length, not an area: geometry supplies A = πr². Units alone cannot determine π or the circular-area relationship.
Cancel
The area unit m2 cancels from m3, leaving m/s. The two 10−6 numerical factors also divide out.
m2 · ms · m2 = m/s
Check
The units are m/s. Scientifically, 0.25 m/s = 250 mm/s; 250 mm/s × 3.14 mm² ≈ 785 mm³/s ≈ 0.8 mL/s, which matches the given flow rate.

Proof It Works

The unit-guided setup agrees with the usual flow-rate equation, after the geometry supplies the area.

Unit Hack setup
m/s = m3/sm2(0.80 × 10⁻⁶ m³/s) ÷ (3.14 × 10⁻⁶ m²)
≈ 0.25 m/s
Conventional equation
v = QA = Qπr2= (0.80 × 10⁻⁶)/(3.14 × 10⁻⁶) m/s
≈ 0.25 m/s

Units can suggest dividing volume per second by area, but they cannot supply the circle-area relationship A = πr² or the coefficient π. Those require geometry and the steady-flow assumptions.

04

Approximately how much mechanical power does the pump deliver against the measured pressure difference?

Not quite. The correct answer is C: 9.6 mW. Pressure difference multiplied by volume flow rate gives hydraulic power under the stated assumptions: 9.6 mW.

Not quite. The correct answer is C: 9.6 mW. Pressure difference multiplied by volume flow rate gives hydraulic power under the stated assumptions: 9.6 mW.

Correct. Pressure difference multiplied by volume flow rate gives hydraulic power under the stated assumptions: 9.6 mW.

Not quite. The correct answer is C: 9.6 mW. Pressure difference multiplied by volume flow rate gives hydraulic power under the stated assumptions: 9.6 mW.

Answer: C. 9.6 mW

Worked solution

Power asks for energy delivered per second. The pressure difference and volume flow rate can form the required units of power without first calculating the trial’s total volume.

Follow the unit setup

Use Next or Previous to review each step.

Target1 / 5
W = Js

Mechanical power is energy transferred per second.

‹
Decode2 / 5
Pa · m3s

Pressure times volume flow rate has power dimensions.

Collect3 / 5
12 kPa→1.2 × 104 Pa
0.80 mLs→0.80 × 10−6 m3s

Convert the supplied pressure and flow rate.

Cancel4 / 5
Nm2 × m2 · ms = Js

Matching area units cancel. The remaining unit is watts.

Check5 / 5
9.6 × 10−3 W = 9.6 mW

Under the given pressure-work assumptions, the result is hydraulic power.

›
12 kPa × 103 Pa1 kPa × 0.80 mLs × 10−6 m31 mL
= 9.6 × 10−3 Pa · m3s = 9.6 × 10−3 W = 9.6 mW

The dimensional cancellation can be made explicit by rewriting pascals as newtons per square meter:

Nm2 × m2 · ms = N · ms = Js = W

Math shortcut: 12 × 0.80 = 9.6 and 103 × 10−6 = 10−3. That is 9.6 milliwatts.

Unit Hack method

Target
Mechanical power, expressed in watts (W).
Decode
A watt is a joule per second. Multiplying pressure by volume flow rate gives the target dimensions.
W = Js = Pa · m3s
Collect
Pressure difference: 12 kPa. Volume flow rate: 0.80 mL/s. Convert kPa to Pa and mL to m3.
Build
Multiply pressure by volume flow rate to construct Pa · m3/s, then check that the pumping conditions make this relationship applicable.
Cancel
kPa and mL cancel their conversion factors. In Pa · m3/s, the m2 factors cancel when Pa is expanded as N/m2, leaving J/s.
Check
The result is 9.6 × 10−3 W. The conventional pressure-work relation confirms that this is the hydraulic power for the assumed constant pressure and flow.

Proof It Works

The unit-guided construction gives the same result as the conventional power equation under the stated steady-flow and negligible-loss assumptions.

Unit Hack setup
J/s = Pa · m³/s(12 kPa)(0.80 mL/s)
9.6 mW
Conventional equation
P = Wt = ΔP · Vt = ΔP · Q= (12 × 10³ Pa)(0.80 × 10⁻⁶ m³/s)
9.6 mW

This relationship comes from pressure work: W = ΔP · V, so power is work per time. Matching dimensions alone would not establish that physical assumption.

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Score: 0 / 41 / 42 / 43 / 44 / 4

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