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R = \(\frac{u^2 sin 2\theta}{g}\) From the above formula, it is evident that range depends on acceleration due to gravity. For real, the range of a projectile is affected by the acceleration due to gravity; it influences how far the projectile travels horizontally.

Given: time of flight T = 4s, g = 10ms\(^{-2}\), H\(_{max}\) \(\theta\) = ? T = \(\frac{2u sin\theta}{g}\) 4 = \(\frac{2 \times u sin \theta}{10}\) 40 = 2u sin\(\theta\) sin\(\theta\) = \(\frac{20}{u}\) But, H\(_{max}\) = \(\frac{u^2 sin^2 \theta}{2g}\) H\(_{max}\) = \(\frac{u^2 \times (\frac{20}{u})^2}{2 \times 10}\) H\(_{max}\) = \(\frac{20 \times 20}{20}\) = 20 metres.(u\(^2\)s cancel out) Thus H\(_{max}\) attained = 20metres.

For the law of conservation of momentum to hold, the system must be closed (no mass enters or leaves) and isolated (no external forces acting on it).

Given: mass of boy = 40 kg = m\(_b\), mass of skateboard = ? = m\(_k\), speed of skateboard v\(_x\) = - 20 m/s, speed of boy v\(_y\) = 1.5 m/s Both the skateboard and the boy were initially at rest. The total momentum before and after the jump must be equal. 0 = m\(_b\) x v\(_y\) + m\(_k\) x v\(_x\) 0 = 40 x 1.5 + m\(_k\) x (-20) 0 = 60 - 20m\(_k\) 20m\(_k\) = 60 m\(_k\) = \(\frac{60}{20}\) = 3 kg.

F\(_c\) = \(\frac{mv^2}{r}\) mass = 200g = 0.2kg, v = 5 m/s, r = 1.5 m F\(_c\) = \(\frac{0.2 \times 5 \times 5}{1.5}\) ≈ 3.33N At the bottom of the circle, the tension T in the string must counteract the weight of the stone as well as provide the centripetal force: T = F\(_c\) + mg T = 3.33 + 0.2 x 10 = 5.3N

F = mass x acceleration due to gravity F\(_c\) = m x a\(_c\) a\(_c\) = a\(_r\) = 4.3 x 10\(^7\)ms\(^{-2}\), m = 3.6 x 10\(^{-6}\)kg F\(_c\) = m x a\(_c\) = 3.6 x 10\(^{-6}\) x 4.3 x 10\(^7\) = 154.8N ≈ 155N ⇒ 1.55 x 10\(^2\)N

Pressure is defined as force per unit area (P = \(\frac{F}{A}\)). When a person stands on one foot, all their weight is concentrated on a smaller area (the sole of one foot), resulting in higher pressure.

Volume of cube = s\(^3\) = 10\(^3\) = 1000cm\(^3\) Volume of cube submerged = 10 cm x 10 cm x 4.5 cm = 450 cm\(^3\) Volume of water displaced = volume of cube submerged = 450 cm\(^3\) Mass of water displaced = 1 cm\(^3\) x 450 cm\(^3\) = 450g (density of water is 1 cm\(^3\)) Mass of cube = V\(_{cube}\) x \(\rho_{cube}\) But, mass of cube = mass of water displaced 1000 cm\(^3\) x \(\rho_{cube}\) = 450g \(\rho_{cube}\) = \(\frac{450}{1000}\) = 0.45 g/cm\(^3\)

The term with a different dimension from the others is D. Torque, as its dimensions ML\(^2\)T\(^{−2}\) which differ from the dimensions of stress and pressure, which are ML\(^{−1}\)T\(^{−2}\)

(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\) FORCE = mass x accel. due to gravity Dimension of force = M.LT\(^{-2}\) Dimension of length = L Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\) (b) The instruments used for measuring the force of gravity at a place are: (1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.

(i) Stress: The base unit of stress is the Pascal (Pa), which is equivalent to kg⋅m\(^{−1}\)⋅s\(^{−2}\) (ii) Force Constant (Spring Constant,k): The base unit of the spring constant is also in Newtons per meter (N/m), which is equivalent to kg⋅s\(^{−2}\) (iii) Planck's Constant (h): The base unit of Planck's constant is Joule seconds (J·s), which is equivalent to kg⋅m\(^2\)s\(^{−1}\) .

ºF = (ºC x 1.8) + 32 But ºC = 60º º F = (60 x 1.8) + 32 = 108 + 32 = 140ºF

As altitude rises, there is less air above a particular point, leading to a decrease in atmospheric pressure. Furthermore, the density of air diminishes with increased altitude, which also contributes to this phenomenon.

Using the equation, R = \(\frac{\rho l}{A}\) we conclude that the wire with the smallest diameter will have the highest resistance i.e. wire E

All the options are reasons why water is not a good thermometric substance, but the most critical one is its lack of uniform expansion. Water's non-linear expansion makes it difficult to correlate temperature changes with volume changes perfectly.