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Generally in thermal/heat conduction, if 1.2 x 106 J is the heat flow per seconds; 30KM-1 is the temperature gradient of the materials; and 400Wm-1K-1 the thermal conductivity of the material, then we have that: heat energy = thermal conductivity x temp. gradient x time x area i.e = 1.2 x 10^6 = 400 x 30 x1x Area => 400 x 30 x Area = 1.2 x 10^6 ∴ Area = (1.2 x 10^6)/(400 x 30) Area = 100m^2 or 1.0 x 10^2m2

From the above diagram, if each of the resistors can dissipate a maximum of 18W, the from the relation power, P = I2R, the current in the 2Ω series resistor is given by: I2 = P/R = 18/2 = 9 => I = √9 = 3A The effective parallel arrangement of the two 2Ω resistors: 1/R = 1/2 + 1/2 = 1Ω ∴ Total resistance in the circuit = 1 + 2 = 3Ω current flowing the circuit = 3A ∴ Maxi. power = P = I2R = 32 x 3 = 27W

In general, if a charged particle is accelerated between two metal plates/a region by a potential difference V volts, the work done on the charge, given by the product of the charge and the potential difference, is equal to the kinetic energy of the charge Thus eV = K.E 1.6 x 10\(^{-19}\) x V = 4.8 x 10\(^{-17}\) ∴ V = \(\frac{4.8 \times 10^{-17}}{1.6 \times 10^{-19}}\) = 300V

A charge in motion always experienced a force on entering a magnetic field

The induced voltage \(\epsilon\) in a coil = N* B* A * \(\omega\) where: N = number of loops (50 loops),B = magnetic field strength (0.4 T),A = area of the loop (0.4 m\(^2\)),\(\omega\) = angular velocity (10 rad/s) \(\epsilon\) in a coil = N* B* A * \(\omega\) = 50 x 0.4 x 0.4 x 10 = 80V Therefore, the amplitude of the induced voltage is 80V

the force on a moving charge in a magnetic field is given by: F = q V β sin θ and this force is maximum when θ = 90o, during which sin 90o = 1: and for sinθ = 0.5, θ = 30o

The stopping potential is the potential required to stop the most energetic photo-emitted electron. From the relation e Vs = K.E,(max) where e is the electron charge, and Vs is the stopping potential => e x Vs = 0.34eV ∴ Vs = (0.34eV) / e = 0.34V ∴ stopping potential = 0.34V

To find the power dissipated in a resistor due to an alternating current (AC), we use the formula for average power: \(P = I_{\text{rms}}^2 R\) where: - \( P \) is the average power, - \( I_{\text{rms}} \) is the root mean square (RMS) value of the current, - \( R \) is the resistance. The current is given by: \(I(t) = 10 \sin(120\pi t)\) The peak current is: \(I_0 = 10 \, \text{A}\) The RMS value of a sinusoidal current is given by: \(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A} \approx 7.07 \, \text{A}\) Using the resistance \( R = 12 \, \Omega \): \(P = I_{\text{rms}}^2 R\) Substituting the values: \(P = (5\sqrt{2})^2 \cdot 12\) Calculating \( (5\sqrt{2})^2 \): \((5\sqrt{2})^2 = 25 \cdot 2 = 50\) Now substituting back into the power formula: \(P = 50 \cdot 12 = 600 \, \text{W}\) Thus, the power dissipated in the resistor is: \(600 \, \text{W}\)

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

R.D = \(\frac{m}{m - m_2} = 5 = \frac{60}{60 - m_1} = \frac{240}{5} = 48N\)

The tendency of a body to remain at rest when a force is applied to it is called inertia. Inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line unless an external force changes that state.

F = \(\frac{mv - mu}{1} = \frac{21 - 16}{0.5} = 10N\)

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

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

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