Question Library

Focal length: \( f = 20 \, \text{cm} \) Object distance: \( u = 100 \, \text{cm} \) (positive since it's in front of the lens) \end{itemize} Using the lens formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) Substituting the known values: \(\frac{1}{20} = \frac{1}{v} + \frac{1}{100}\) Rearranging to solve for \( v \): \(\frac{1}{v} = \frac{1}{20} - \frac{1}{100}\) Finding a common denominator (100): \(\frac{1}{20} = \frac{5}{100}\) Thus, \(\frac{1}{v} = \frac{5}{100} - \frac{1}{100} = \frac{4}{100}\) Solving for \( v \): \(v = \frac{100}{4}\) = 25 cm

F= 15cm M = 3 M = \(\frac{v}{u}\) V =3u \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\) \(\frac{1}{15}\) = \(\frac{1}{u}\) - \(\frac{1}{3u}\) \(\frac{1}{15}\) = \(\frac{2}{3u}\) U =10cm D = u + v = 10 +30 = 40cm

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

\(\frac{R_1}{R_2}\) = \(\frac{(100- L_1)}{( L_1)}\) \(\frac{R_1}{2}\) = \(\frac{45}{55}\) R\(_1\) = 2 \times \(\frac{45}{55}\) = 1.64\Omega this is because the known resistor is connected to 0.00cm end and balances at 55cm

Power drawn by the resistor = IV = (2)(3) = 24W = 24J/s Energy dissipated = power \times time = (24)(5)(60) = 7200J

Weight in Air = 45N Weight in Water = 15N R.D = \(\frac{45}{45-15}\) = 1.50

2 \times Distance = speed \times time 2 \times Distance = (340) \times (0.8) = 272/2 Distance = 136m

N = \(\frac{1}{sin\theta}\) = \(\frac{1}{sin34}\) = 1.79

Every form of energy is measured in joules (J).

Elastic modulus is the ratio of pressure (stress) applied to a body to the resistance (strain) produced by the body. = \(\frac{stress}{strain}\)

The height of the mercury column in a simple barometer is calculated as the atmospheric pressure.

Inertia is the resistance of the object to any change in its motion, including a change in direction. An object will stay still or keep moving at the same speed and in a straight line, unless it is acted upon by an external unbalanced force.

In SHM, the body must have acceleration in a direction opposite to the displacement, and the acceleration must be directly proportional to the displacement. Also, the acceleration must be directed towards its mean position. This is not the case with a football on the pitch. Also, in SHM, the restoring force must act towards the equilibrium point.

The pressure of the water is much greater deeper down and the dam needs to be thick at the bottom so that it is strong enough to withstand this larger pressure.

Total force exerted on the block = (10 - 1)N =9N Force = mass x acceleration 9 = m x 1.5 m = 9/1.5 = 6kg