This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here's the breakdown of your questions: 1. When a sphere is released to fall through a column of water, the following forces act on it: • Weight: The gravitational force acting downwards, due to the mass of the sphere. • Upthrust (Buoyant Force): The upward force exerted by the fluid, equal to the weight of the water displaced by the sphere. • Drag Force (Fluid Resistance/Viscous Force): The force opposing the motion of the sphere through the water, acting upwards. 2. Given: Side of cube, L = 0.3 cm = 0.003 m Density of wood, _wood = 700 kg/m^3 Density of water, _water = 1000 kg/m^3 Step 1: Determine the submerged height of the cube. For a floating object, the weight of the object equals the upthrust. Weight of cube = _wood V_total g = _wood L^3 g Upthrust = _water V_submerged g = _water (L^2 h_sub) g Equating the forces: _wood L^3 g = _water L^2 h_sub g _wood L = _water h_sub h_sub = _wood_water L Substitute the values: h_sub = 700 kg/m^31000 kg/m^3 × 0.003 m h_sub = 0.7 × 0.003 m h_sub = 0.0021 m Step 2: Calculate the distance from the top of the cube to the water surface. This is the portion of the cube that is above the water. Distance above water = Total height of cube - Submerged height d_above = L - h_sub d_above = 0.003 m - 0.0021 m d_above = 0.0009 m Converting to centimeters: d_above = 0.0009 m × 100 cm/m d_above = 0.09 cm The distance from the top of the cube to the water surface is 0.09 cm. 2. (b) Two applications of the induction coil are: • Ignition systems in internal combustion engines (e.g., spark plugs in cars). • X-ray tubes and other high-voltage discharge tubes. 3. (i) Given: Capacitance, C = 1.20 F Resistance, R = 300 RMS voltage, V_rms = 20 V Frequency, f = 1000 Hz Circuit diagram for a capacitor and a non-inductive resistance connected in series across an AC supply: (0,0) to[R=R] (2,0) to[C=C] (4,0); (4,0) -- (4.5,0); (0,0) -- (-0.5,0); (-0.5,0) to[sV=V_rms] (-0.5,2) -- (4.5,2) -- (4.5,0); (Note: The diagram above is a textual representation. In a proper drawing, the AC source would be a circle with a sine wave symbol inside, connected to the series R-C components.) 4. Given: Charge of electron, q = 1.6 × 10^-19 C Magnetic field strength, B = 0.8 Wb/m^2 = 0.8 T Speed of electron, v = 2.9 × 10^7 m/s Mass of electron, m_e = 9.11 × 10^-31 kg 4. (i) Radius, r, of the orbit Step 1: Equate the magnetic force to the centripetal force. When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force for circular motion. F_B = F_c qvB = (m_e v^2)/(r) Step 2: Solve for the radius r. r = (m_e v)/(qB) Substitute the given values: r = (9.11 × 10^-31 kg)(2.9 × 10^7 m/s)(1.6 × 10^-19 C)(0.8 T) r = 2.6419 × 10^-231.28 × 10^-19 m r ≈ 2.064 × 10^-4 m The radius of the orbit is 2.064 × 10^-4 m. 4. (ii) Centripetal force experienced by the charge Step 1: Use the formula for magnetic force (which is equal to centripetal force). F_c = qvB Substitute the given values: F_c = (1.6 × 10^-19 C)(2.9 × 10^7 m/s)(0.8 T) F_c = 3.712 × 10^-12 N The centripetal force experienced by the charge is 3.712 × 10^-12 N. 5. Angular magnification (or magnifying power) as used in relation to a telescope in normal adjustment is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye. For a telescope in normal adjustment, the final image is formed at infinity. It can also be expressed as the ratio of the focal length of the objective lens to the focal length of the eyepiece. 5. (i) Given: Distance between lenses, L = 82.5 cm Angular magnification, M = 10 5. (i) (a) Name the type of thin lens used The type of thin lens used in an astronomical telescope is a converging lens (or convex lens) for both the objective and the eyepiece. 5. (i) (b) Calculate the focal lengths of each lens Step 1: Set up equations based on the given information. For an astronomical telescope in normal adjustment: 1) Angular magnification: M = (f_o)/(f_e) 2) Distance between lenses: L = f_o + f_e Where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece. Step 2: Substitute the given values into the equations. From (1): 10 = (f_o)/(f_e) f_o = 10 f_e From (2): 82.5 cm = f_o + f_e Step 3: Solve the system of equations. Substitute f_o = 10 f_e into the second equation: 82.5 = 10 f_e + f_e 82.5 = 11 f_e f_e = (82.5)/(11) f_e = 7.5 cm Now find f_o: f_o = 10 f_e = 10 × 7.5 cm f_o = 75 cm The focal length of the objective lens is 75 cm and the focal length of the eyepiece is 7.5 cm. 5. (ii) 5. (ii) (i) One difference between circular and rotational motion • Circular motion describes the movement of a point mass or a particle along a circular path, where the object's center of mass traces a circle. Rotational motion describes the movement of a rigid body about an internal axis, where the entire body spins. 5. (ii) (ii) Two conditions for which the acceleration of a body can be zero The acceleration of a body is zero when: • The body is at rest (its velocity is zero and constant). • The body is moving with a constant velocity (constant speed in a straight line). That's 2 down. 3 left today — send the next one.