Here are the full solutions for each question:

1. Force Exerted by a Hydraulic Cylinder

Given:

• Diameter = 10 cm = 0.1 m
• Radius $r = \frac{0.1}{2} = 0.05$ m
• Pressure $P = 700$ kPa = $700,000$ Pa

Area of the cylinder: $A = \pi r^2 = \pi (0.05)^2 = \pi (0.0025) \approx 0.00785 \, \text{m}^2$

Force exerted: $F = P \times A = 700,000 \times 0.00785 \approx 5493.15 \, \text{N}$

2. Force Produced by a Hydraulic Press

Given:

• Diameter = 0.3 m
• Radius $r = \frac{0.3}{2} = 0.15$ m
• Pressure $P = 500$ kPa = $500,000$ Pa

Area of the piston: $A = \pi r^2 = \pi (0.15)^2 = \pi (0.0225) \approx 0.07069 \, \text{m}^2$

Force produced: $F = P \times A = 500,000 \times 0.07069 \approx 35344.8 \, \text{N}$

3. Pressure in the Hydraulic System

Given:

• Force $F = 4000$ N
• Area $A = 0.05$ m²

Pressure: $P = \frac{F}{A} = \frac{4000}{0.05} = 80000 \, \text{Pa} = 80 \, \text{kPa}$

4. Pressure from a Hydraulic Press

Given:

• Force $F = 12000$ N
• Area $A = 0.1$ m²

Pressure: $P = \frac{F}{A} = \frac{12000}{0.1} = 120000 \, \text{Pa} = 120 \, \text{kPa}$

5. Work Done by a Hydraulic Lift

Given:

• Force $F = 1500$ N
• Distance $d = 3$ m

Work done: $W = F \times d = 1500 \times 3 = 4500 \, \text{J}$

6. Work Done Moving a Load

Given:

• Force $F = 2000$ N
• Distance $d = 2.5$ m

Work done: $W = F \times d = 2000 \times 2.5 = 5000 \, \text{J}$

7. Energy Stored in a Hydraulic System

Given:

• Pressure $P = 600$ kPa = $600,000$ Pa
• Volume $V = 0.3$ m³

Energy stored: $E = P \times V = 600,000 \times 0.3 = 180000 \, \text{J}$

8. Energy Stored in a Hydraulic Tank

Given:

• Pressure $P = 400$ kPa = $400,000$ Pa
• Volume $V = 0.4$ m³

Energy stored: $E = P \times V = 400,000 \times 0.4 = 160000 \, \text{J}$

9. Power Output of a Hydraulic Pump

Given:

• Work $W = 10000$ J
• Time $t = 20$ s

Power: $P = \frac{W}{t} = \frac{10000}{20} = 500 \, \text{W}$

10. Power of a Hydraulic System

Given:

• Energy $E = 5000$ J
• Time $t = 10$ s

Power: $P = \frac{E}{t} = \frac{5000}{10} = 500 \, \text{W}$

11. Torque Produced by a Hydraulic Motor

Given:

• Force $F = 400$ N
• Radius $r = 0.25$ m

Torque: $\tau = F \times r = 400 \times 0.25 = 100 \, \text{Nm}$

12. Torque from a Hydraulic Wrench

Given:

• Force $F = 600$ N
• Distance $d = 0.3$ m

Torque: $\tau = F \times d = 600 \times 0.3 = 180 \, \text{Nm}$

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If you have any questions or need further details on any of these calculations, feel free to ask! Here are some related questions to consider:

1. How would changes in pressure affect the force exerted by the hydraulic cylinder?
2. What is the relationship between work and energy in hydraulic systems?
3. How does torque affect the performance of hydraulic motors?
4. What are the efficiency factors to consider in hydraulic systems?
5. How can you increase the power output of a hydraulic system?

Tip: Always double-check the units when calculating in hydraulic systems to ensure consistency!