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To find the point of intersection, equate the two equations ⇒ 2x\(^2\) + 10 = 4x + 16 ⇒ 2x\(^2\) + 10 - 4x - 16 = 0 ⇒ 2x\(^2\) - 4x - 6 = 0 Factorize ⇒ 2(x + 1)(x - 3) = 0 ∴ x = -1 or 3 The curves will intersect at -1 and 3 Area = \(\int^b_a\) [upper function] - [lower function] dx = \(\int^3_1 4x + 16 - (2x^2 + 10) dx\) = \(\int^3_1 -2x^2 + 4x + 6dx\) = \((\frac {-2x^3}{3} + 2x^2 + 6x)^3_1\) = \((\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))\) = 18 - \((\frac {10}{3})\) = 18 + \(\frac {10}{3}\) = \(\frac {64}{3}\)

The patient cannot see clearly an object closer than 50 cm Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens u=25cm ;v=-50cm (virtual image); p= \(\frac{1}{f}\)=\(\frac{1}{u}\) + \(\frac{1}{v}\) ⇒\(\frac{1}{f}\) =\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = \(\frac{1}{25}\)-\(\frac{1}{50}\) ⇒\(\frac{1}{f}\) = f = 50cm =0.5m ⇒\(\frac{1}{f}\) = \(\frac{2-1}{50}\) p = \(\frac{1}{f}\) p = \(\frac{1}{0.5}\) p = 2 diopters ; The patient needs a converging lens with a power of 2 diopters

Mechanics is the branch of physics that deals with the motion of objects and the forces acting on them. It is the foundation of all other branches of physics, and it is used to study everything from the motion of planets to the motion of electrons.

m=1200kg, \(V_1\)= \(10ms^{-1}\) \(V_2\) = \(15ms^{-1}\), w= ? work=►K.E = \( K.E_2\) = \(K.E_2\) - \(K.E_1\) ⇒work= \(\frac{1}{2}{mv^2_2}-\frac{1}{2}{mv^2_1}\) ⇒work= \(\frac{1}{2}m({v^2_2}-{v^2_1}\)) ⇒work= \(\frac{1}{2}\times 1200\times (15^2-10^2)\) ⇒work = 600 \times (225 -100) ⇒work= 600 \times 125 ⇒work= 7.5\times\(10^4 j\)

The work done against the gravitational force is calculated using the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately \(9.8 ms^2\) on Earth), and h is the height. Substituting the given values, we get W = 3.0 kg * \(9.8 m/s^2\) * 240 m = 7.2 kJ. Therefore, the correct answer is '7.2 kJ'.

ε= 80%, L=200\times 10 = 2000N, \(d_L\)=10m \(d_E\)= 110-10 = 100m, E=? ε = \(\frac{work done on the load}{work done by the effort}\) \times 100% ⇒ work done on load = 2000 \times10 =20,000 j ⇒ work done by effort = E \times 100 = 100E ⇒ 80 = \(\frac{20,000}{100E}\) \times 100% ⇒ 80 = \(\frac{20,000}{E}\) ⇒ E = \(\frac{20,000}{80}\) ⇒ E = 250N

We first find the lorry's initial velocity a = \(4ms^-2\), s= 250m, t= 10, u=? ⇒ ut + \(\frac{1}{2}\times4\times(10^2) ⇒250= 10u + 200 ⇒ 250-200 = 10u ⇒ 50 = 10u ⇒U =\frac{50}{10}\) = \(5ms^-1\) For the next 10s, we set t=20s = s = (5 \times 20) + \(\frac{1}{2}\times4\times{20^2}\) = s = 100 + 800 ⇒ 900 m ;Distance covered:900-250 =650m

W = mg = 220 x 9.8 = 2156 N ⇒Sin 38º = \(\frac{2156}{T_1}\) ⇒ \(T_1\) = \(\frac{2156}{Sin 38}\) ⇒ \(T_1\) = 3502 N Cos 38º = \(\frac{T_2}{T_1}\) ⇒ \(T_2\) = 3502 x Cos 38º ⇒ \(T_2\) = 2760 N ; \(T_1\) = 3502 N, \(T_2\) = 2760 N.

Fx= 12cos32^\circ-16cos26^\circ = -4.20N (right is taken +ve and left is taken -ve) Fy=12sin32^\circ+16sin26^\circ-21 = -7.63N (up is taken +ve and down is taken -ve) \(R^2\)= 4.202+7.632 ⇒\(R^2\) = 75.86 ⇒ R=\sqrt75.86 ∴R = 8.71 N tan \theta=\(\frac{opp}{adj}\) ⇒tan \theta=\(\frac{7.63}{4.20}\) ⇒tan \theta =1.8095 ⇒ \theta = \(tan^-1\) (1.8095) \theta = 61^\circ ∴The angle of the resultant force with the x-axis is 61^\circ

W = mg = 0.5 x 10 = 5 N Since it's light, neglect the weight of the metre rule. The effective tension T acting in the vertical direction = T sin 30^\circ From the second condition of equilibrium, sum of clockwise moments equal sum of anticlockwise moments Taking moment at E ⇒ T sin 30^\circ x 30 = 5 x 50 ⇒ ∴T=250 T = \(\frac{250}{15}\) = 16.67N

The working of the beam balance is based on the principle of moments. The principle of moments states that the sum of the clockwise moments on a beam must be equal to the sum of the counterclockwise moments on the beam. This means that the weight of the object on the left side of the beam must be multiplied by the distance from the fulcrum to the left side of the beam, and the weight of the object on the right side of the beam must be multiplied by the distance from the fulcrum to the right side of the beam. If these two products are equal, then the beam will be in equilibrium.

A couple is a pair of forces that are equal in magnitude but opposite in direction, and that are applied to a body at different points. The forces of a couple do not produce any translation, but they do produce a rotation.

U= 75 \(ms^-1\) =22º, g= 10 \(ms^-2\) , R=? R= \(\frac{U^2sin2\theta}{g}\) ⇒R= \(\frac{75^2\timessin2(22)}{10}\) ⇒R= \(\frac{5625\timesSin44}{10}\) ⇒R= \(\frac{5625\times0.6947}{10}\) ⇒R= \(\frac{3907.69}{10}\) ⇒ 391m

Object A will undergo unstable equilibrium as it can move farther from its original position Object B will undergo stable equilibrium as it can return to its original position Object C will undergo neutral equilibrium as it can remain in a new position

m=2kg, r=1.2m, f=120rev/min ; T=? f= \(\frac{120rev}{1min}\) × \(\frac{1min}{60s}\) = 2rev\s w=2πf= 2×π×2= 4πrad\s T= \(\frac{mv^2}{r}\), but v = wr ⇒ T = \(\frac{(m)(wr)^2}{r}\) ⇒ T = \(mw^2r\) = T = 2× (4π)^2 × 1.2 T = 379N ( to 3 s.f)