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The effective force pushing it forward is the horizontal component of the force = F\(_x\) = F Cos\(\theta\) F\(_x\) = F Cos\(\theta\) = 50 x Cos 60º = 50 x 0.5 = 25N

Efficiency = \(\frac{\text{ M.A}}{\text{V.R}}\) x 100 M.A = \(\frac{\text{LOAD}}{\text{EFFORT}}\) V.R = 2, Eff. = 60% = 0.6, Load = 300N 0.6 = \(\frac{\text{M.A}}{2}\) M.A = 1.2 1.2 = \(\frac{300}{\text{E}}\) Effort = \(\frac{300}{1.2}\) = 250N

The velocity ratio of a machine is 4 simply means that the distance moved by effort is 4 times the distance moved by the load.

Mechanical Advantage = \(\frac{\text{load}}{\text{effort}}\) Given: mass(m) = 60kg, acceleration due to gravity(g) = 10ms\(^2\) and effort = 40N Load = m x g = 60 x 10 = 600N M.A = \(\frac{600}{40}\) = 15

Velocity ratio of an inclined plane = \(\frac{1}{Sin\theta}\) = \(\frac{1}{Sin 60}\) ≈ 1.155

Acceleration is defined as the rate of change of velocity. This means a bus can accelerate by changing its speed, direction, or both. However, the most general definition of acceleration involves a change in velocity.

If a body in linear motion changes from point P to Q, the motion is Translational. Translational motion refers to the movement of an object from one location to another in a straight line, without any rotation involved.

the car is moving with uniform acceleration.

Mean acceleration = \(\frac{\Delta V}{\Delta t}\) Point 1: (2s, 5m/s) Point 2: (4s,15m/s) Mean acceleration = \(\frac{ 15 - 5 }{4 - 2}\) = 5.0ms\(^2\)

Distance = Speed \timesTime Distance = 335 x 5 = 1675m

The man's weight = T - ma mass = 60kg, a = 3ms\(^2\) The man's weight = m(g - a) = 60(10 - 3) = 60 x 7 = 420N

The attraction between molecules of the same substance is referred to as cohesive force. These forces enable molecules to adhere to one another, which affects properties such as surface tension in liquids and the structural integrity of materials.

Density = \(\frac{\text{mass}}{\text{volume}}\) Volume of solid cube = s\(^3\) = 1.5\(^3\) = 3.375cm\(^3\) Density = 2700kgm\(^{-3}\) ⇒ 2.7gcm\(^{-3}\) 2.7 = \(\frac{\text{m}}{3.375}\) mass of the cube = 2.7 x 3.375 = 9.1125g \(\approxeq\) 9.1g

Density = \(\frac{\text{mass}}{\text{volume}}\) mass of hydrometer = y kg 80% hydrometer's volume submerge(20% above the fluid) = \(\frac{4}{5}\) x 2y x 10\(^{-5}\) = \(\frac{8y \times 10^{-5}}{5}\) Density = y x \(\frac{5}{8y\times 10^{-5}}\) = 6.25 x 10\(^4\)kgm\(^3\)

For Hare's apparatus Relative density = \(\frac{h_w}{h_l}\) Given: height of liquid = 0.5cm, height of water = 0.3cm Relative density = \(\frac{0.3}{0.5}\) = 0.6