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King Saud University College of Science 14 marks Department of Physics and Astronomy Second Term (1430-31) Thursday 13 Jumada II 1431 h 1st mid-term 27 May 2010 Phys 145 (106) 10 30 a.m Q1 A box is pulled on horizontal non friction surface by a rope that makes a 60° angle to the horizontal direction. The tension in the rope is T = 70 N, and the box moves a distance d = 15 m. The work done in pulling the box is: a. 288 J b. 250 J c. 125 J d. 525 J e. 500 J 0 W= T× cos60 × d = 70 × 0.5 ×15 = 525 J Q2 A 1500 kg car moving at 10 m/s hits head-on a 3000 kg in rest. If the collision is completely inelastic, the dissipated mechanical energy is a. 25 kJ K1 = b. 100 kJ c. 150 kJ e. 50 kJ 1 1 m1v12 = × 1500 × 10 2 = 75000 J 2 2 m1v1 = (m1 + m2 )V ⇒ V = K' = d. 75 kJ m1v1 = (m1 + m2 ) m12 v12 1 1 1 m12 v12 1 (1500) 2 × 10 2 (m1 + m 2 )V 2 = (m1 + m2 ) = = × = 25000 J 2 2 4500 (m1 + m2 ) 2 2 (m1 + m 2 ) 2 ∆K = K 1 − K ' = 75000 − 25000 = 50000J = 50 kJ Q3. A pipe of radius “r” ends with a nozzle of radius “r/3”. If the average velocity of water flowing in the pipe is v1, the average velocity v2 of water leaving the nozzle is: v v a. v1 b. 1 c. 3v1 d. 1 e. 9 v1 9 3 Apply the equation of continuity in the form A1 v1 = A2 v2 Where v1 = v (the average velocity in the pipe) and v2 (the average velocity in the nozzle) =? A1 = π (r)2 A2 = π (r/3)2 Thus π (r)2 v = π (r/3)2 v2 = 1/9 π (r)2 v2 v2 = 9 v Q4. Two circular metal plates capacitor with radius 0.2 m are separated in vacuum by 10-3 m. The capacitance of this parallel plates capacitor is: (ε0=8.85×10-12 C2N-1m-2) b. 1.1x10-9F c. 8.85x10-11F e. None of the above a. 2.8x10-10F d. 3,55 x10-10F C= (ε0A where A= πr2 , =[(8.85x10-12 C2N-1m-2)3.14x(0.2m)2] 10-3m =1.1x10-9F Q5 The number of electrons flowing through a battery that delivers a current of 5.0 A for 12s is: a. 4 b. 3.75 × 1020 c. 7.83 × 1015 d. 6.54 × 1013 e. 56 Q6 A room air conditioner uses 500 W of electrical power. If it operates during 20 hours per day and the energy costs 8 Halala per kilowatt hour, the daily operating cost is: b. 1 SR c. 8 SR d. 0.8 S R e. 1.6 S R a) 10 SR The daily operating cost= (0.5kW)(20 h)(0.08 R/kwh)= 0.8 SR Q1 Q2 Q3 Q4 Q5 Q6 d e e b b d Q9 -A 4.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 9 N. Find the speed of the block after it has moved 5.0 m. F ∆x Solution The work done by this force is W=F∆ ∆x = (9 N) (5.0 m) = 45 J ( 0.75 point ) Using the work-kinetic energy theorem and noting that the initial kinetic energy is zero, we obtain: ( 0.75 point ) ( 0.5 point ) Q10 In the following distribution of charges, the electric field and potential at point P midway between the two fixed charges are respectively: Solution 3.375x103 NC-1 -3x10-6 C P + ( 0.75 point ) E= k[ = +3x10-6 4m , ( 0.25 point ) Q11 and are Two resistors arranged in a circuit that carries a total current of 9 A as shown in the figure. What are the current through and the voltage across ? 4m ( 0.75 point ) Vp= k[ = 0 volt ( 0.25 point ) I=9 A R2 Solution The equivalent resistor R for the two parallel resistors R1=2Ω and R2=4Ω is (0.75 point) The voltage across the resistor R1 is the same as that across R and therefore (0.75 point) The current through the resistor R2 is (0.5 point)