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Upthrust = weight fluid displacement 0.040 - 0.024 = PVg 0.016 = 800 x V x 10 V = 2.0 x 106

The swing rate of a pendulum is determined by its length. The longer the pendulum, the slower the pendulum swings and vice-versa.

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 E.

F = m(v - u)/t; (v - u) = Ft/m = 8.0

μ= F/mg = 20/40 = 0.5

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

v2 = u2 + 2gh = 0 + 2 x 10 x 10 v2 = 200; K.E = 1/2mv2 = 1/2 x 5 x 200 = 500J

Gap ΔL = αL1Δθ = 1.0 x 10-5 x 3 x (40 - 28) = 3.6 x 10-2

Under standard atmospheric pressure, pure water freezes at 0ºC. Water is most dense at around 4ºC. As it cools below this temperature, it begins to expand.

Using the genaral gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) T\(_1\) = 30º = 273 + 30 = 303k T\(_2\) = ? let V\(_1\) = V, V\(_2\) = \(\frac{2}{3}\)V, P\(_1\) = P, P\(_2\) = 2P From, \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) T\(_2\) = \(\frac{P_2 V_2 T_1}{P_1 V_1}\) = \(\frac{2P \times \frac{2}{3}V \times 303}{P \times V}\) = 404k(the Ps and Vs cancel out) 404k → ºC = 404 - 273 = 131ºC

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

Using Charles law: V1/T1 = V2/T2; (283 * 303) / 283 V2= 303

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

Pressure at a depth for fluid with constant density such as mercury is given as \(p = p_{0} + \rho hg\) where \(p_{0}\) = atmospheric pressure. \(p_{0} = \rho hg =13600 \times \frac{76}{100} \times 10\) = \(103,360 Nm^{-2}\) \(p = 103,360 + (13600 \times \frac{10}{100} \times 10)\) \(p = 103,360 + 13,600\) \(p = 116,960 Nm^{-2} \approxeq 1.17 \times 10^{5} Nm^{-2}\)