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Magnetic poles always exist in pairs. It is not possible to produce just an isolated north pole or an isolated south pole. If a bar magnet is broken into two pieces, each piece will have a north and south pole.

For a car battery rated 45 AH, being charged by a charger of 2.5 A rating, the time for the battery to be fully charged is given by: T = (45)/2.5 = 18 hrs

The electric force experienced in an electric field of intensity E is given by: F = qE: where q = the test charge. E = F/Q = 0.4/10-10 = 0.4 X 10-10 = 4.0 X 109 NC-1

The charge must be opposite in sign, since the positive charges will deflect to the negative plate, and the negative charge to the positive plate

To determine the deceleration of the motorcyclist, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance: \(v^2 = u^2 + 2as\) Where: \( v \) = final velocity (0 m/s, since he stops at the traffic light), \( u \) = initial velocity (30 m/s), \( a \) = acceleration (deceleration, which will be negative) \( s \) = distance (50 m) Substituting the values into the equation, we rearrange it to solve for \( a \): \(0 = (30)^2 + 2a(50)\) Calculating \( (30)^2 \): \(0 = 900 + 100a\) Now, rearranging for \( a \): \(100a = -900\) \(a = -\frac{900}{100} = -9 \, \text{m/s}^2\) The deceleration of the motorcyclist is \( \mathbf{9 \, m/s^2} \).

For the two point charges, the magnitude of the electrostatic force between them is given by the equation: F ∝ 1/r2, and if the distance is increased by a factor of 4, that is new distance = 4r, then, new force F ∝ 1/(4r)2 = ∝ 1/16r2 Thus it will be 1/16 of the former value

The fig above shows a conventional transistor, showing the three basic features, the Base, the Emitter and the collector. Thus comparing this with the figure in the question it implies that P = Emitter, Q = Base and R = collector

\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2^1_0X \) In the reaction above, balancing the left with the right shows that X is a neutron

Generally, from Hares Experiment, density is inversely proportional to height i.e h ∝1/d; → h1d1 = h2d2 ∴ 8 x 103 = h2 x 800 ∴ h2 = 8 x 103_800 = 10cm

From the heat exchange involved, the latent heat of vaporization, L can be deduced from the relation: ML = 106, where M = mass of the liquid. ∴ 2 x L = 106 => L = 106/2 = 5.0 x 105J kg-1

The effective capacitance in the circuit is given by: 1/C = 1/2 + 1/3 + 1/6 = 6/6 therefore C = 1μF or 1.0 x 10-6F. Since C = Q/V, its implies that the charges Q, flowing in the circuit is given by Q = CV = 1.0 x 10-6 x 12 = 12.0 x 10-6 Coulomb. ∴ P.d across the 6μF, V = Q/C = 12.0 x 10-66.0 x 10-6F = 2.0 volts

From the given graph, when the angle of minimum deviation = 18º, the angle of incidence = 25º

F = \(\frac{17}{sin 60} = \frac{17}{0.866}\) = 20N

Normal reaction R = 3 kg x 10 = 30N coefficient of friction , μ = 0.5 ∴ acting frictional force = μR = 30 x 0.5 Pulling/sliding force F = 6kg x 10 = 60N. ∴ Net downward force = ma = 60 - 15.0 => (3 + 6)a = 45 ∴ 9a = 45 a = 5m/s2

If the string is in equilibrium, the the tension in the string Q = 4N + 5N = 9N