最新人教版高中物理必修2课后习题答案名师优秀教案.doc
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1、人教版高中物理必修2课后习题答案Fifth chapters First curve motion 1. answer: as shown in Figure 6-12, in the A and C positions, the head speed is the same as the speed of V entering the water; the speed of the head in the position of B and D is opposite to the speed of V when entering water. 2. answer: car driving
2、half a week, the speed change direction of 180 degrees. Each car driving speed 10s, changing the direction of 30 degrees, the velocity vector diagram as shown in figure 613. 3. answer: as shown in Figure 6-14, AB segment is the curve movement, BC segment and CD segment is linear motion curve motion.
3、 The motion of second particles in a plane 1. solution: the projectile in the horizontal direction of the sub speed is VX = 800 x cos60 degrees = 400m/s; shells in the vertical direction of the sub speed is vy = 800 x sin60 degrees = 692m/s. Figure 6-15. 2.: according to the problem, when there is n
4、o jumper with the speed of V2, the effect of wind speed to the east to enable him to obtain V1, V as the landing speed speed of V2 and V1, as shown in Fig. 6-15, and vertical angle theta theta, Tan = 0.8, theta = 38.7 DEG 3. answer: should be West side. As shown in Figure 6-16, because have the same
5、 boat and shells from west to East speed V1, hit the target rate of V is V1 and fired V2 speed speed, so the speed of V2 should be fired by some. 4. answer: as shown in figure 617. Third, the law of projectile motion 1. solution: (1) the motorcycle can cross the ditch. The horizontal displacement of
6、 the motorcycle is y, = 1.5m = experience time, displacement in horizontal direction = VT = x = 40 * 0.55M = 22m 20m, so the motorcycle can cross the ditch. In general, when flying in the air, the front wheels are always higher than the rear wheels. When landing, the rear wheels are landing first. (
7、2) motorcycle landing in vertical direction of the velocity angle of vy = GT = 9.8 = 5.39m/s * 0.55m/s motorcycle landed in the horizontal direction and the vertical direction of the speed of the speed of VX = v = 40m/s motorcycle landing speed of motorcycle landing of the theta theta = VX, Tan / vy
8、 = 405.39 = 7.42 2. solution: the car has been speeding. Parts of horizontal projectile motion in the vertical direction, the displacement of y = 2.45m = time, in the horizontal displacement of x = VT = 13.3m, the initial velocity components of horizontal projectile motion is: v = x / T = 13.3 / 0.7
9、1m/s = 18.7m/s = 67.4km/h 60km/h so the car has been speeding. Answer: (1) let the ball from a slope position A no initial release; horizontal distance measurement x ball on the ground point P and the edge of the table; the vertical distance measurement of Y ball on the ground with the ball resting
10、at the level of placement P desktop center. The initial speed of the ball leaving the table is. The fourth experiment is to study the horizontal throwing motion 1. answer: also need equipment is scale. Experimental procedure: (1) Regulation of wood height, the surface and the ball on the board when
11、the center of the left level desktop distance for a certain value of y; (2) let the ball from a slope on a certain position, A without initial release; (3) measure the distance between the falling point of the ball on the board and the distance between the P1 and the heavy vertical line x1; (4) Regu
12、lation of wood height, the surface and the ball on the board when the center of the left level desktop distance for a certain value of 4y; (5) release the ball from the same position on the slope without releasing the initial velocity of the A; (6) measure the distance between the falling point of t
13、he ball on the board and the distance between the P2 and the heavy vertical line x2; (7) compare X1 and X2, and if 2x1 = X2, it shows that the ball moves uniformly in horizontal direction. Change the distance between the wall and the heavy vertical, x, measure the vertical distance between the point
14、 of fall and the throw point y. If 2x1 = X2, 4Y1 = Y2, it shows that the ball moves uniformly in horizontal direction. Fifth circular motion 1. solutions: the two objects located at the equator and Beijing are equal in angular velocity with the rotation of the earth, and are equal to the angular vel
15、ocity of the circle. The object at the equator is moving at a uniform speed along the rotation of the earth, V1 = Omega R = 465.28m/s, and the angular velocity of V2 is equal to omega Rcos40 DEG = 356.43m/s at the same speed as the rotation of the earth on the earths rotation in Beijing Solution: 2.
16、 minute cycle for T1 = 1H, T2 = 12h clockwise cycle (1) the angular velocity of minute hand and hand the ratio of Omega 1, Omega 2 = T2: T1 = 12: 1 (2 minute) tip and tip clockwise line speed ratio is V1: 1r1: V2 = Omega Omega 2r2 = 14.4: 1 3. answer: (1) A and B, two line speed is equal, angular ve
17、locity and radius is inversely proportional (2) A and C have the same angular velocity at two points, and the line velocity is proportional to the radius (3) the radius of two points of B and C is equal, and the line velocity is in direct proportion to the angular velocity Explanation: the purpose o
18、f this question is to let students understand the relationship between line speed, angular speed and radius: v = Omega R; at the same time understand the physical meaning that the driving device does not skid is that the line speed between the contact points is equal. 4. need to measure the big, sma
19、ll gear and the rear wheel radius R1, R2, r3. The speed of the bike Explanation: the purpose of this topic is to enable students to understand the relationship between uniform speed, circular motion, and linear speed, angular velocity, and radius between them. However, the motion of any point on the
20、 wheel is not circular motion, and its trajectory is the roller line. Therefore, in dealing with this problem, the wheel should be taken as the reference object, and the ground is in contact with the wheel without skidding, so that the speed of the ground moving to the right is equal to the line spe
21、ed at the back of the rear wheel. 5. solution: disk rotation cycle of T = 0.2S (1) scan the time of each sector t = T/18 = 1/90s. (2) the number of bytes in each sector is 512, and the number of bytes read in 1s is 90 * 512 = 46080. Explanation: the purpose of this topic is to let students understan
22、d the uniform circular motion in light of the actual situation. Sixth quarter centripetal acceleration 1. answer: A. A and B line speed is equal to the use of small radius of the centripetal acceleration. Therefore, the centripetal acceleration of B is large; when the B. and B cycles are equal, the
23、radial acceleration of the radius is large. Therefore, the centripetal acceleration of a nail is great; When the angular velocity of C. and B is equal, the radial acceleration is large with an = V omega. Therefore, the centripetal acceleration of B is small; When the velocities of D. and B are equal
24、, the angular acceleration is large with an = V omega. Since the angle between the center of a circle and the center of a circle is equal to that of B in equal time, the angular velocity of a horn is large, and the centripetal acceleration of a body is large. Explanation: the purpose of this subject
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