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Given: - Load \( W = 180 \, \text{N} \) - Efficiency \( \eta = 90\% = 0.9 \) - Let the effort \( E \) be applied. The efficiency of a machine is given by the formula: \(\eta = \frac{\text{Mechanical Advantage (MA)}}{\text{Velocity Ratio (VR)}}\) Mechanical Advantage (MA) is defined as the ratio of load to effort: \(\text{MA} = \frac{W}{E}\) Determine the Velocity Ratio (VR) Given that the effort arm is twice the load arm, we can express the Velocity Ratio (VR) as: \(\text{VR} = \frac{\text{Length of effort arm}}{\text{Length of load arm}} = \frac{2L}{L} = 2\) Substitute into the efficiency formula Substituting the expressions for MA and VR into the efficiency formula gives: \(0.9 = \frac{\frac{W}{E}}{2}\) Rearrange to find the effort \( E \) Multiplying both sides by \( 2E \): \(1.8E = W\) Substituting \( W = 180 \, \text{N} \): \( 1.8E = 180\) Solve for \( E \) Now, dividing by 1.8: \(E = \frac{180}{1.8} = 100 \, \text{N}\) Therefore, the effort applied by the machine is: \(\boxed{100 \, \text{N}}\).

The three 2µf capacitors from the left are parallel. Therefore, effective capacitance in parallel = 6µf The 6µf, 2µf(opposite the 3µf), and the 3µf are in series. So effective/total capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) + \(\frac{1}{C_3}\) = \(\frac{1}{6}\) + \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{1 + 3 + 2}{6}\) = \(\frac{6}{6}\) C\(_T\) = 1µf

The last postulate of the kinetic molecular theory states that the average kinetic energy of a gas particle depends only on the temperature of the gas. Thus, the average kinetic energy of the gas particles increases as the gas becomes warmer.

An impulse is equal to the net force on the object times the time period over which this force is applied. Impulse is derive from the equation F = ma, which comes from Newton's second law of motion

Diffusion is the net movement of molecules/atoms from a region of high concentration to low concentration. Brownian motion is the erratic random movement of microscopic particles in a field. Diffraction is the various phenomena that occur when a wave encounters an obstacle.

Volume of liquid displaced = \(\frac{2}{3}\)(0.5)\(^3\) = 0.0833cm3 Mass of liquid displaced = mass of floating cube = 75kg Density of liquid = \(\frac{mass}{volume}\) = \(\frac{75}{0.0833}\) = 900kgm\(^{-3}\) R.D of liquid = \(\frac{(900 )}{(1000)}\) = 0.9

Recall V = F \times \(\lambda\) = 260 \times 1.29 = 335.4ms\(^{-1}\) Echo time = \(\frac{2}{2}\) = 1 sec Distance = v x t = 335.4 x 1 = 335.4m

Fuse wires have a low melting point so that if a huge amount of current is passed through it to any appliance than that wire would melt and stop the appliance from getting damaged by not letting the current pass any further.

P = 40W R = 90W V = ? P = \(\frac{V^2}{R}\) V\(^2\) = PR = 40 x 90 V\(^2\) = 3000 V = \sqrt3600 V = 60N

In case of small drops of mercury, the gravitational potential energy is negligible in comparison to the potential energy due to surface tension.Consequently, to keep the drop in equilibrium, the mercury drop’s surface tends to contract so that its surface area will be the least for a sphere and the drops will be spherical. But in the case of bigger drops of mercury, the potential energy due to gravity is predominant over the potential energy due to surface tension.Consequently, to keep equilibrium , the mercury drop tends to assume minimum potential energy as possible, the drop becomes oval in shape and lower center of gravity.

Water 1 => M = 150g = 0.15kg, \theta\(_2\) = 60\(^o\)C; \theta\(_1\) = \theta Q\(_1\) = MC (\theta\(_2\) – \theta\(_1\)) = 0.15C (60 - \theta) Water 2 => M = 300g = 0.3kg \theta1 = 20\(^o\)C, \theta\(_2\) = \theta Q2 = MC (\theta\(_2\) – \theta\(_1\)) = 0.3C (\theta – 20) Combining Q1 and Q2 0.15C (60 - \theta) = 0.3C (\theta – 20) 9 – 0.15\theta = 0.3\theta – 6 0.3\theta + 0.15\theta = 9 + 6 0.45 \theta = 15 \theta = 33.33\(^o\)C

Brownian motion, also called Brownian movement, any of various physical phenomena in which some quantity is constantly undergoing small, random fluctuations. If a number of particles subject to Brownian motion are present in a given medium and there is no preferred direction for the random oscillations, then over a period of time the particles will tend to be spread evenly throughout the medium.

using the formular : Moment = Force \times Distance 5 X 0 = 0Nm

X = \(\frac{4}{3}\) r = ? Shell’s law:. = \(\frac{Sin i^0}{ Sin r^0}\) = \(\frac{Sin 30^0}{Sin r^0}\) Sin\(r^0\) = \(\frac{3 Sin 30^0}{4}\) Sin \(r^0\) = 0.375 R \(^o\) = Sin-1 (0.375) R \(^o\) = 22.02 \(^o\) R \(^o\) = 22 \(^o\)