170 likes | 319 Views
Physics (H). 3 rd Quarter Exam Review. 3 rd Quarter Exam Review. An ideal gas confined in a box initially has pressure p. If the absolute temperature
E N D
Physics (H) 3rd Quarter Exam Review
3rd Quarter Exam Review • An ideal gas confined in a box initially has pressure p. If the absolute temperature • of the gas is doubled and the volume of the box is quadrupled, the pressure is (A) p/8 (B) p/4 (C) p/2 (D) p (E) 2p C) p/2 2. A mass m of helium gas is in a container of constant volume V. It is initially at pressure p and absolute (Kelvin) temperature T. Additional helium is added, bringing the total mass of helium gas to 3m. After this addition, the temperature is found to be 2T. What is the gas pressure? (A) 2/3 p (B) 3/2 p (C) 2 p (D) 3 p (E) 6P E) 6P
3rd Quarter Exam Review • 3. A monatomic ideal gas at pressure P = 105 Pa is in a container • of volume V = 12 m3 while at temperature T = 50ºC. How many • molecules of gas are in the container? • (A) 447 (B) 2888 (C) 6.02 × 1023 (D) 2.83 × 1026 (E) 1.74 × 1027 D) 2.83 x 10 26 4. A steel wire, 150m in length at 10 oC, has a coefficient of linear expansion of 11 x 10 –6 /C o. Give its change in length as the temperature increases to 45 oC. a. 0.65 cm b. 1.8 cm c. 5.8 cm d. 12 cm C) 5.8 cm
3rd Quarter Exam Review 5. One mole of an ideal gas has a temperature of 100°C. If this gas fills the 10.0m3 volume of a closed container, what is the pressure of the gas? (A) 0.821 Pa (B) 3.06 Pa (C) 83.1 Pa (D) 310 Pa (E) 1.84 × 1024 Pa (D) 310 Pa 6. What is the Kelvin scale temperature that is equivalent to 10 degrees on the Fahrenheit scale? (A) 283 K (B) 273 K (C) 261 K (D) 10 K (E) 10 K (C) 261 K
3rd Quarter Exam Review 7. An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended into heat for warming water. How much mechanical energy is required to increase the temperature of 300. g of water from 20 C to 95 C? c = 4186 J/kg-C a. 94000 J b. 22000 J c. 5400 J d. 14 J a. 94000 J 8. A 2.00 kg copper rod is 50.00 cm long at 23 C. If 40000 J are transferred to the rod by heat, what is its changed in length? c = 387 J/kg-C, = 17 x 10 –6/ C a. 0.022 cm b. 0.044 cm c. 0.059 cm d. more information is needed b. 0.044 cm
3rd Quarter Exam Review 9. Iced tea is made by adding ice to 1.8 kg of hot tea at 80 C. How many kg of ice at 0 C are required to bring the mixture to 10 C? Lf = 3.33 x 10 5 J/kg-C, c = 4186 J/kg- C a. 1.8 kg b. 1.6 kg c. 1.4 kg d. 1.2 kg c. 1.4 kg 10. How much energy must be removed from 100 g of oxygen at 22 C to liquefy it at –183 C? c = 912.5 J/kg-C, Lv = 2.13 x 10 5J/kg a. 57348J b. 40007 J c. 21903 J d. 9042 J b. 40007 J
3rd Quarter Exam Review 11. The period of a spring-mass system undergoing simple harmonic motion is T. If the amplitude of the spring-mass system’s motion is doubled, the period will be:(A) ¼ T (B) ½ T (C) T (D) 2T (E) 4T (C) T 12. A particle oscillates up and down in simple harmonic motion. Its height y as a function of time t is shown in the diagram. At what time t does the particle achieve its maximum positive acceleration?(A) 1 s (B) 2 s (C) 3 s (D) 4 s E) None of the above (A) 1 s
3rd Quarter Exam Review 13. Two objects of equal mass hang from independent springs of unequal spring constant and oscillate up and down. The spring of greater spring constant must have the (A) smaller amplitude of oscillation (B) larger amplitude of oscillation (C) shorter period of oscillation (D) longer period of oscillation (E) lower frequency of oscillation (A) smaller amplitude of oscillation 14. String L and string H have the same tension and length. String L has mass m and string H has mass 4m. If the speed of the waves in string L is v, the speed of the waves in string H isA) v/2 B) v C) 1.4 v D) 2v E) 4v A) v/2
3rd Quarter Exam Review 15. Assume that waves are propagating in a uniform medium. If the frequency of the wave source doubles then A) The speed of the waves doubles B) the wavelength do the waves doubles C) the speed of the waves halves D) the wavelength of the waves halves E) none of the above D) the wavelength of the waves halves . 16. What is the electric force of two electrons 10 –10 m apart? a. 2.3 x 10 –8 N b. 4.6 x 10 –8 N c. 9.2 x 10 –8 N d. 2.3 x 10 –9 N e. 4.6 x 10 –10 N a. 2.3 x 10 –8 N
3rd Quarter Exam Review 17. The electric field halfway between two positive point charges of 1.7 x 10 –5 C is: a. 34 N/C b. 17 N/C c. 0 N/C d. –17 N/C e. –34 N/C c. 0 N/C • 18. What is the electric field halfway between two collinear negative charges • separated by 6 cm. One charge is – 2.1 C, whereas the other is – 1.3 C. • 2.6 x 10 5 N/C b. 6.3 x 10 5 N/C c. 3.2 x 10 6 N/C • d. 8.0 x 10 6 N/C e. 1.1 x 10 7 N/C d. 8.0 x 10 6 N/C
3rd Quarter Exam Review 19. A positive charge of 10–6 coulomb is placed on an insulated solid conducting sphere. Which of the following is true?(A) The charge resides uniformly throughout the sphere.(B) The electric field inside the sphere is constant in magnitude, but not zero.(C) The electric field in the region surrounding the sphere increases with increasing distance from the sphere.(D) An insulated metal object acquires a net positive charge when brought near to, but not in contact with, the sphere.(E) When a second conducting sphere is connected by a conducting wire to the first sphere, charge is transferred until the electric potentials of the two spheres are equal.
3rd Quarter Exam Review 20. The following configurations of electric charges are located at the vertices of an equilateral triangle. Point P is equidistant from the charges. In which configuration is the electric field at P equal to zero? (A)A (B) B (C) C (D) D (E) E
3rd Quarter Exam Review Free Response 1.Two point charges, Q1 and Q2, are located a distance 0.20 meter apart, as shown above. Charge Q1 = +8.0C. The net electric field is zero at point P, located 0.40 meter from Q1 and 0.20 meter from Q2. a. Determine the magnitude and sign of charge Q2. b. Determine the magnitude and direction of the net force on charge Q1 c. Determine the coordinate of the point R on the x–axis between the two charges at which the electric field is zero.
3rd Quarter Exam Review • E = kQ/r2 and since the field is zero E1 + E2 = 0 giving k(Q1/r12 + Q2/r22) = 0This gives the magnitude of Q2 = Q1(r22/r12) = 2C and since the fields must point in opposite directions from each charge at point P, Q2 must be negative. • b. F = kQ1Q2/r2 = 3.6 N to the right (they attract) • c. between the charges we have a distance from Q1 of x and from Q2 of (0.2 m – x)E = kQ1/x + kQ2/(0.2 m – x) = 0, solving for x gives x = 0.16 m
3rd Quarter Exam Review 2. A charge Q1 = –1.6 x 10–6 coulomb is fixed on the x–axis at +4.0 meters, and a charge Q2 = + 9 x 10–6 coulomb is fixed on the y–axis at +3.0 meters, as shown on the diagram above. a. Calculate the magnitude of the electric field E1 at the origin O due to charge Q1 b. Calculate the magnitude of the electric field E2 at the origin O due to charge Q2. a. E = kq/r2 = 900 N/C b. E = kq/r2 = 9000 N/C
3rd Quarter Exam Review 3. An ideal spring of unstretched length 0.20 m is placed horizontally on a frictionless table as shown above. One end of the spring is fixed and the other end is attached to a block of mass M = 8.0 kg. The 8.0 kg block is also attached to a mass less string that passes over a small frictionless pulley. A block of mass m = 4.0 kg hangs from the other end of the string. When this spring-and-blocks system is in equilibrium, the length of the spring is 0.25 m and the 4.0 kg block is 0.70 m above the floor. (a) Calculate the tension in the string. (b) Calculate the force constant of the spring. The string is now cut at point P. (c) Calculate the time taken by the 4.0 kg block to hit the floor. (d) Calculate the frequency of oscillation of the 8.0 kg block. (e) Calculate the maximum speed attained by the 8.0 kg block
3rd Quarter Exam Review (a) Calculate the tension in the string. (b) Calculate the force constant of the spring. The string is now cut at point P. (c) Calculate the time taken by the 4.0 kg block to hit the floor. (d) Calculate the frequency of oscillation of the 8.0 kg block. (e) Calculate the maximum speed attained by the 8.0 kg block a) Simply isolating the 4 kg mass at rest. Fnet = 0 Ft – mg = 0 Ft = 39 N b) Tension in the string is uniform throughout, now looking at the 8 kg mass, Fsp = Ft = k∆x 39 = k (0.05) k = 780 N/m d) 4 kg mass is in free fall. D = vit + ½ g t2 – 0.7 = 0 + ½ (– 9.8)t2 t = 0.38 sec e) First find the period. then the frequency is given by f = 1/T = 1.6 Hz The 8 kg block will be pulled towards the wall and will reach a maximum speed when it passes the relaxed length of the spring. At this point all of the initial stored potential energy is converted to kinetic energy Usp = K ½ k ∆x2 = ½ mv2 ½ (780) (0.05)2 = ½ (8) v2 v = 0.49 m/s