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The term Adhesive forces refers to the attractive forces between unlike substance

Momentum = Mass x Velocity Change in Momentum = Mass x (change in velocity) Change in Momentum = Mass x (v - u) Mass = 100g = 0.1kg U = 2ms\(^{-1}\) V = 10ms\(^{-1}\) Change in momentum = m (v − u) = 0.1 (2 -(- 10)) = 0.1 (12) = 1.2Ns

V. R = \(\frac{1}{Sin\theta}\) m = 20kg V.R = \(\frac{1}{Sin 30}\) = 2 Efficiency = 75% Load = mg = 20 x 10 = 200N Efficiency = \(\frac{M.A}{V.R} \times 100\) 75/100 = \(\frac{M.A}{2}\) M .A = \(\frac{75 \times 2}{100}\) M. A = 1.5 Since M. A. = \(\frac{\text{Load}}{\text{Force}}\) Force = \(\frac{200}{1.5}\) = 133.3N

Inertia, property of a body by virtue of which it opposes any agency that attempts to put it in motion or, if it is moving, to change the magnitude or direction of its velocity. Inertia is a passive property and does not enable a body to do anything except oppose such active agents as forces and torques. A moving body keeps moving not because of its inertia but only because of the absence of a force to slow it down, change its course, or speed it up. There are two numerical measures of the inertia of a body: its mass, which governs its resistance to the action of a force, and its moment of inertia about a specified axis, which measures its resistance to the action of a torque about the same axis.

\(\frac{I}{V}\) = \(\frac{1}{f}\) – \(\frac{1}{u}\) \(\frac{I}{V}\) = \(\frac{1}{10}\) – \(\frac{1}{20}\) = \(\frac{1}{20}\) V V = 20cm. M = \(\frac{V}{U}\) = \(\frac{20}{20}\) = 1

The energy needed to move a unit positive charge around a complete electric circuit is indeed referred to as Electromotive Force (EMF).

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option D.

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