A simple pendulum 4 m long swings with an amplitude of 0.2m. Compute the velocity of the pendulum at its lower point and its acceleration at extreme end.

You're on a roll — 14. A simple pendulum 4 m long swings with an amplitude of 0.2m. Compute the velocity of the pendulum at its lower point and its acceleration at extreme end. Given: Length of pendulum, L = 4 m Amplitude, A = 0.2 m Acceleration due to gravity, g = 9.8 m/s^2 i) Velocity at the lower point (v_max): Step 1: Calculate the angular frequency . = sqrt((g)/(L)) = sqrt(9.8 m/s)^24 m = sqrt(2.45 s)^-2 ≈ 1.565 rad/s Step 2: Calculate the maximum velocity (v_max) at the lower point. v_max = A v_max = (0.2 m)(1.565 rad/s) v_max ≈ 0.313 m/s The velocity of the pendulum at its lower point is 0.313 m/s. ii) Acceleration at the extreme end (a_max): Step 1: Calculate the maximum acceleration (a_max) at the extreme end. a_max = A^2 a_max = (0.2 m)(1.565 rad/s)^2 a_max = (0.2 m)(2.45 s^-2) a_max = 0.49 m/s^2 The acceleration of the pendulum at the extreme end is 0.49 m/s^2. 17. The earth's total magnetic field intensity at a place is 4.5 × 10^-7 T and angle of dip is 30^. Calculate horizontal and vertical components. Given: Total magnetic field intensity, B = 4.5 × 10^-7 T Angle of dip, = 30^ i) Horizontal component (B_H): Step 1: Use the formula for the horizontal component. B_H = B Step 2: Substitute the given values. B_H = (4.5 × 10^-7 T) (30^) B_H = (4.5 × 10^-7 T) (sqrt(3)2) B_H = (4.5 × 10^-7 T)(0.866) B_H ≈ 3.897 × 10^-7 T The horizontal component is 3.90 × 10^-7 T. ii) Vertical component (B_V): Step 1: Use the formula for the vertical component. B_V = B Step 2: Substitute the given values. B_V = (4.5 × 10^-7 T) (30^) B_V = (4.5 × 10^-7 T)(0.5) B_V = 2.25 × 10^-7 T The vertical component is 2.25 × 10^-7 T. 18. An object is placed 15cm from a converging lens of focal length 10cm. Find the image distance and magnification. Given: Object distance, u = -15 cm (real object, placed to the left of the lens) Focal length of converging lens, f = +10 cm i) Image distance (v): Step 1: Use the lens formula. (1)/(f) = (1)/(v) - (1)/(u) Step 2: Rearrange to solve for v. (1)/(v) = (1)/(f) + (1)/(u) Step 3: Substitute the given values. (1)/(v) = (1)/(10 cm) + (1)/(-15 cm) (1)/(v) = (1)/(10 cm) - (1)/(15 cm) Step 4: Find a common denominator and simplify. (1)/(v) = (3)/(30 cm) - (2)/(30 cm) (1)/(v) = (1)/(30 cm) v = 30 cm The image distance is 30 cm. (Positive sign indicates a real image formed on the opposite side of the lens). ii) Magnification (M): Step 1: Use the magnification formula for a lens. M = (v)/(u) Step 2: Substitute the calculated image distance and given object distance. M = 30 cm-15 cm M = -2 The magnification is -2. (Negative sign indicates an inverted image, and |M|=2 means it is twice the size of the object). 19. A bar magnet of magnetic length 10cm has a magnetic moment of 1.2 Am^2. Calculate the magnetic field intensity at a point 20cm from each pole. Given: Magnetic length, 2l = 10 cm = 0.1 m So, half magnetic length, l = 5 cm = 0.05 m Magnetic moment, M = 1.2 Am^2 Distance from each pole to the point, r = 20 cm = 0.2 m This point is on the equatorial line (perpendicular bisector) of the magnet, as it is equidistant from both poles. Step 1: Use the formula for the magnetic field on the equatorial line of a bar magnet. B_eq = (_0)/(4) (M)/((r^2)^3/2) Since r is the distance from each pole to the point, and r^2 = x^2 + l^2 where x is the distance from the center of the magnet to the point, the formula simplifies to: B_eq = (_0)/(4) (M)/(r^3) Step 2: Substitute the known values. The constant (_0)/(4) = 10^-7 T m/A. B_eq = (10^-7 T m/A) 1.2 Am^2(0.2 m)^3 B_eq = (10^-7 T m/A) 1.2 Am^20.008 m^3 B_eq = (10^-7 T m/A) (150 A/m) B_eq = 1.5 × 10^-5 T The magnetic field intensity at the given point is 1.5 × 10^-5 T.