To find the angle $\theta$, use the sine function:

sin⁡(θ)=OppositeHypotenuse=725\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7}{25}sin(θ)=HypotenuseOpposite​=257​

Calculate $\theta$:

θ=sin⁡−1(725)≈25.8∘\theta = \sin^{-1}\left(\frac{7}{25}\right) \approx 25.8^\circθ=sin−1(257​)≈25.8∘

Thus, the angle $\theta$ is approximately 25.8 degrees.

First, calculate the area of the cylinder:

Area=π(d2)2=π(52)2=π×6.25≈19.63 in2\text{Area} = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{5}{2}\right)^2 = \pi \times 6.25 \approx 19.63 \text{ in}^2Area=π(2d​)2=π(25​)2=π×6.25≈19.63 in2

The force $F$ is then:

F=Pressure×Area=1500 PSI×19.63 in2≈7850 lbsF = \text{Pressure} \times \text{Area} = 1500 \text{ PSI} \times 19.63 \text{ in}^2 \approx 7850 \text{ lbs}F=Pressure×Area=1500 PSI×19.63 in2≈7850 lbs

Thus, the force exerted by the cylinder is approximately 7850 lbs.

The surface area $A$ of a cylinder is given by:

$$A=2\pi r(r+h)$$

where $r$ is the radius and $h$ is the height. Substituting $r=3\text{ cm}$ and $h=10\text{ cm}$:

A=2π×3×(3+10)=2π×3×13=2π×39≈1236 cm2A = 2 \pi \times 3 \times (3 + 10) = 2 \pi \times 3 \times 13 = 2 \pi \times 39 \approx 1236 \text{ cm}^2A=2π×3×(3+10)=2π×3×13=2π×39≈1236 cm2

Thus, the surface area of the cylinder is approximately 1236 cm².

Calculate the total volume pumped by:

$$\text{Total Volume}=\text{Flow Rate}\times \text{Time}$$

Convert 3 hours into minutes:

$$\text{Time}=3\text{ hours}\times 60\text{ minutes/hour}=180\text{ minutes}$$

Thus:

$$\text{Total Volume}=20\text{ liters/minute}\times 180\text{ minutes}=3600\text{ liters}$$

Thus, the total fluid pumped is 3600 liters.

Frictional force FfF_fFf​ is calculated using:

Ff=μ×NF_f = \mu \times NFf​=μ×N

where $\mu$ is the coefficient of friction (0.3) and $N$ is the normal force. The normal force $N$ is equal to the weight of the block:

N=m×g=50 kg×10 m/s2=500 NN = m \times g = 50 \text{ kg} \times 10 \text{ m/s}^2 = 500 \text{ N}N=m×g=50 kg×10 m/s2=500 N

Thus:

Ff=0.3×500 N=150 NF_f = 0.3 \times 500 \text{ N} = 150 \text{ N}Ff​=0.3×500 N=150 N

Thus, the frictional force is 150 N.

To find the final amount $A$, use the formula:

A=P⋅ertA = P \cdot e^{rt}A=P⋅ert

where $P=1000$, $r=0.05$, and $t=3$. Substituting these values:

A=1000⋅e0.05×3A=1000⋅e0.15e0.15≈1.1618A=1000⋅1.1618=1157.63A = 1000 \cdot e^{0.05 \times 3} \\ A = 1000 \cdot e^{0.15} \\ e^{0.15} \approx 1.1618 \\ A = 1000 \cdot 1.1618 = 1157.63A=1000⋅e0.05×3A=1000⋅e0.15e0.15≈1.1618A=1000⋅1.1618=1157.63

Thus, the final amount $A$ is approximately $1157.63.

The rotational speed of gears in a meshing system is inversely proportional to their number of teeth. Using the formula:

Speed of Gear B=Speed of Gear A×Number of Teeth in Gear ANumber of Teeth in Gear B\text{Speed of Gear B} = \text{Speed of Gear A} \times \frac{\text{Number of Teeth in Gear A}}{\text{Number of Teeth in Gear B}}Speed of Gear B=Speed of Gear A×Number of Teeth in Gear BNumber of Teeth in Gear A​

Substituting the values:

Speed of Gear B=120×2040=120×0.5=60 RPM\text{Speed of Gear B} = 120 \times \frac{20}{40} = 120 \times 0.5 = 60 \text{ RPM}Speed of Gear B=120×4020​=120×0.5=60 RPM

Thus, the rotational speed of Gear B is 60 RPM.

Use the Law of Sines to solve for side $b$:

asin⁡(A)=bsin⁡(B)\frac{a}{\sin(A)} = \frac{b}{\sin(B)}sin(A)a​=sin(B)b​

Given A=30∘A = 30^\circA=30∘, B=45∘B = 45^\circB=45∘, and $a=10$ cm:

10sin⁡(30∘)=bsin⁡(45∘)100.5=b0.707120=b0.7071b=20×0.7071=14.2 cm\frac{10}{\sin(30^\circ)} = \frac{b}{\sin(45^\circ)} \\ \frac{10}{0.5} = \frac{b}{0.7071} \\ 20 = \frac{b}{0.7071} \\ b = 20 \times 0.7071 = 14.2 \text{ cm}sin(30∘)10​=sin(45∘)b​0.510​=0.7071b​20=0.7071b​b=20×0.7071=14.2 cm

Thus, the length of side $b$ is approximately 14.2 cm.

To meet the required pressure, the valve must be adjusted:

$$\text{Required Adjustment}=\text{Required Pressure}-\text{Current Pressure}\text{Required Adjustment}=3000\text{ PSI}-2500\text{ PSI}=500\text{ PSI}$$

Thus, the valve should be increased by 500 PSI to meet the required pressure.

The volume $V$ of a cone is given by:

V=13πr2hV = \frac{1}{3} \pi r^2 hV=31​πr2h

where $r$ is the radius (4 cm) and $h$ is the height (9 cm). Substituting the values:

V=13π×42×9V=13π×16×9V=13π×144V≈13×452.4=113.1 cm3V = \frac{1}{3} \pi \times 4^2 \times 9 \\ V = \frac{1}{3} \pi \times 16 \times 9 \\ V = \frac{1}{3} \pi \times 144 \\ V \approx \frac{1}{3} \times 452.4 = 113.1 \text{ cm}^3V=31​π×42×9V=31​π×16×9V=31​π×144V≈31​×452.4=113.1 cm3

Thus, the volume of the cone is approximately 113.1 cm³.

Pascal’s Principle states that the pressure applied to a fluid is transmitted equally throughout the fluid. Pressure $P$ is:

P=FAP = \frac{F}{A}P=AF​

For the small piston:

P=100 N0.01 m2=10000 PaP = \frac{100 \text{ N}}{0.01 \text{ m}^2} = 10000 \text{ Pa}P=0.01 m2100 N​=10000 Pa

For the larger piston:

F=P×A=10000 Pa×0.2 m2=2000 NF = P \times A = 10000 \text{ Pa} \times 0.2 \text{ m}^2 = 2000 \text{ N}F=P×A=10000 Pa×0.2 m2=2000 N

Thus, the force exerted by the larger piston is 1000 N.

Work $W$ is calculated by:

$$W=F\times d$$

where $F$ is the force (weight of the load) and $d$ is the distance (height). Rearranging to solve for $F$:

F=Wd=5000 J10 m=500 NF = \frac{W}{d} = \frac{5000 \text{ J}}{10 \text{ m}} = 500 \text{ N}F=dW​=10 m5000 J​=500 N

Thus, the weight of the load is 1000 N.

The surface area $A$ of a cylinder is given by:

$$A=2\pi r(r+h)$$

where $r$ is the radius (5 cm) and $h$ is the height (12 cm). Substituting these values:

A=2π×5×(5+12)A=2π×5×17A=170π≈170×3.1416A≈1068.58 cm2A = 2 \pi \times 5 \times (5 + 12) \\ A = 2 \pi \times 5 \times 17 \\ A = 170 \pi \approx 170 \times 3.1416 \\ A \approx 1068.58 \text{ cm}^2A=2π×5×(5+12)A=2π×5×17A=170π≈170×3.1416A≈1068.58 cm2

Thus, the surface area is approximately 1,070 cm².

Ohm’s Law states that:

$$V=I\times R$$

where $V$ is the voltage (24 V), $I$ is the current, and $R$ is the resistance (8 ohms). Rearranging to solve for $I$:

I=VR=24 V8 ohms=3 AI = \frac{V}{R} = \frac{24 \text{ V}}{8 \text{ ohms}} = 3 \text{ A}I=RV​=8 ohms24 V​=3 A

Thus, the current flowing through the circuit is 3 A.

The mechanical advantage (MA) of a pulley system is given by:

MA=Load ForceEffort Force\text{MA} = \frac{\text{Load Force}}{\text{Effort Force}}MA=Effort ForceLoad Force​

where the load force is 1000 N and the effort force is 200 N:

MA=1000 N200 N=5\text{MA} = \frac{1000 \text{ N}}{200 \text{ N}} = 5MA=200 N1000 N​=5

Thus, the mechanical advantage of the pulley system is 5.

To find the actual dimensions, multiply the blueprint dimensions by the scale factor:

$$\text{Actual Length}=\text{Blueprint Length}\times \text{Scale Factor}\text{Actual Length}=4\text{ cm}\times 50=200\text{ cm}=2\text{ m}\text{Actual Width}=5\text{ cm}\times 50=250\text{ cm}=2.5\text{ m}$$

Thus, the actual dimensions are 8 m by 10 m.

Pressure $P$ is given by:

P=FAP = \frac{F}{A}P=AF​

where $F$ is the force (150 N) and $A$ is the area (0.02 m²). Substituting these values:

P=150 N0.02 m2=7500 PaP = \frac{150 \text{ N}}{0.02 \text{ m}^2} = 7500 \text{ Pa}P=0.02 m2150 N​=7500 Pa

Thus, the pressure applied to the fluid is 7,500 Pa.

The temperature should be maintained between 70°C and 80°C. Given the current temperature is 65°C, the temperature needs to be increased to fall within the optimal range:

$$\text{Required Increase}=70\text{°C}-65\text{°C}=5\text{°C}$$

Thus, to bring the temperature into the optimal range, increase it by at least 10°C.

Work $W$ is given by:

$$W=F\times d$$

where $F$ is the force (150 N) and $d$ is the distance. Rearranging to solve for $d$:

d=WF=600 J150 N=4 md = \frac{W}{F} = \frac{600 \text{ J}}{150 \text{ N}} = 4 \text{ m}d=FW​=150 N600 J​=4 m

Thus, the distance over which the object was moved is 4 m.

Use the trigonometric function cosine for this problem:

$$\text{Height}=\text{Ladder Length}\times \sin (\text{Angle})$$

where the ladder length is 10 meters and the angle is 60°:

Height=10×sin⁡(60∘)Height=10×0.866≈8.66 m\text{Height} = 10 \times \sin(60^\circ) \\ \text{Height} = 10 \times 0.866 \approx 8.66 \text{ m}Height=10×sin(60∘)Height=10×0.866≈8.66 m

Thus, the ladder reaches approximately 8.66 meters up the wall.

The mechanical advantage (MA) of a lever is given by:

MA=Effort Arm LengthLoad Arm Length\text{MA} = \frac{\text{Effort Arm Length}}{\text{Load Arm Length}}MA=Load Arm LengthEffort Arm Length​ MA=41=4\text{MA} = \frac{4}{1} = 4MA=14​=4

The load force can be calculated by multiplying the effort force by the MA:

$$\text{Load Force}=\text{Effort Force}\times \text{MA}\text{Load Force}=200\text{ N}\times 4=800\text{ N}$$

Thus, the load that can be lifted is 800 N.

First, calculate the total resistance RtotalR_{\text{total}}Rtotal​ in the series circuit:

Rtotal=10 Ω+20 Ω+30 Ω=60 ΩR_{\text{total}} = 10 \text{ Ω} + 20 \text{ Ω} + 30 \text{ Ω} = 60 \text{ Ω}Rtotal​=10 Ω+20 Ω+30 Ω=60 Ω

Using Ohm’s Law to find the current:

I=VRtotalI=30 V60 Ω=0.5 AI = \frac{V}{R_{\text{total}}} \\ I = \frac{30 \text{ V}}{60 \text{ Ω}} = 0.5 \text{ A}I=Rtotal​V​I=60 Ω30 V​=0.5 A

Thus, the current flowing through the circuit is 0.5 A.

According to Pascal’s Principle, the pressure is the same throughout the fluid:

Pressure=ForceArea\text{Pressure} = \frac{\text{Force}}{\text{Area}}Pressure=AreaForce​

For the small piston:

Pressure=100 N0.01 m2=10,000 Pa\text{Pressure} = \frac{100 \text{ N}}{0.01 \text{ m}^2} = 10,000 \text{ Pa}Pressure=0.01 m2100 N​=10,000 Pa

For the large piston:

Forcelarge=Pressure×ArealargeForcelarge=10,000 Pa×0.5 m2=5,000 N\text{Force}_{\text{large}} = \text{Pressure} \times \text{Area}_{\text{large}} \\ \text{Force}_{\text{large}} = 10,000 \text{ Pa} \times 0.5 \text{ m}^2 = 5,000 \text{ N}Forcelarge​=Pressure×Arealarge​Forcelarge​=10,000 Pa×0.5 m2=5,000 N

Thus, the force exerted by the large piston is 5,000 N.

The noise level exceeds the maximum allowable limit of 85 dB. According to safety guidelines, exceeding noise limits requires implementing noise reduction measures to protect workers’ hearing and ensure compliance with safety regulations.

The ratio given is 3:2:1. For every 2 parts of sand, there are 3 parts of cement. To find the amount of cement needed:

Cement=(32)×120 kg of sand=180 kg of cement\text{Cement} = \left(\frac{3}{2}\right) \times 120 \text{ kg of sand} = 180 \text{ kg of cement}Cement=(23​)×120 kg of sand=180 kg of cement

However, since the ratio is maintained and scaled, for the given sand, the amount of cement needed is 90 kg.

The force required to push the object up the plane includes both the force to overcome gravity and the force to overcome friction. The gravitational force component acting down the plane is:

Fgravity=mgsin⁡(θ)=200×9.8×sin⁡(30∘)=980 NF_{\text{gravity}} = mg \sin(\theta) = 200 \times 9.8 \times \sin(30^\circ) = 980 \text{ N}Fgravity​=mgsin(θ)=200×9.8×sin(30∘)=980 N

The frictional force is:

Ffriction=μ×mgcos⁡(θ)=0.2×200×9.8×cos⁡(30∘)=336 NF_{\text{friction}} = \mu \times mg \cos(\theta) = 0.2 \times 200 \times 9.8 \times \cos(30^\circ) = 336 \text{ N}Ffriction​=μ×mgcos(θ)=0.2×200×9.8×cos(30∘)=336 N

Total force required is:

Ftotal=Fgravity+Ffriction=980 N+336 N=1316 NF_{\text{total}} = F_{\text{gravity}} + F_{\text{friction}} = 980 \text{ N} + 336 \text{ N} = 1316 \text{ N}Ftotal​=Fgravity​+Ffriction​=980 N+336 N=1316 N

The minimum force required is approximately 784 N.

For resistors in parallel, the voltage across each resistor is the same. The current through each resistor is given by Ohm’s law I=VRI = \frac{V}{R}I=RV​:

For the 4 Ω resistor:

I1=12 V4 Ω=3 AI_1 = \frac{12 \text{ V}}{4 \text{ Ω}} = 3 \text{ A}I1​=4 Ω12 V​=3 A

For the 6 Ω resistor:

I2=12 V6 Ω=2 AI_2 = \frac{12 \text{ V}}{6 \text{ Ω}} = 2 \text{ A}I2​=6 Ω12 V​=2 A

The total current supplied by the power supply is the sum of the currents:

Itotal=I1+I2=3 A+2 A=5 AI_{\text{total}} = I_1 + I_2 = 3 \text{ A} + 2 \text{ A} = 5 \text{ A}Itotal​=I1​+I2​=3 A+2 A=5 A

The diagonal of a rectangle can be found using the Pythagorean theorem:

d=a2+b2=502+1202=2500+14400=16900=130 metersd = \sqrt{a^2 + b^2} = \sqrt{50^2 + 120^2} = \sqrt{2500 + 14400} = \sqrt{16900} = 130 \text{ meters}d=a2+b2​=502+1202​=2500+14400​=16900​=130 meters

Thus, the length of the diagonal path is 130 meters.

Operating a hydraulic pump above its maximum temperature can lead to system failure, overheating, or damage to components. The best course of action, as stated in the manual, is to shut down the pump to prevent any potential hazards and allow it to cool before resuming operation.

Safety regulations require that elevator cars be level with the floor to prevent tripping hazards and ensure safe entry and exit for passengers. Even a small discrepancy like 2 cm should be corrected immediately. Failure to do so could result in serious injury, especially for individuals with mobility issues.