![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in Figure 1, two long enough straight metal guides Mn and PQ are placed in parallel. The distance between the two guides is l0, m and P, and there is a resistance of R indirectly. A uniform straight metal rod AB with mass of M is placed on the two guides and perpendicular to the guide rail. The whole device is in a uniform magnetic field with magnetic induction intensity of B, and the magnetic field direction is perpendicular to the inclined plane downward. The resistance of the guide rail and the metal rod can be ignored Let the AB rod slide down along the guide rail from rest, and the guide rail and the metal rod have good contact, regardless of the friction between them. (1) the device seen from B to a direction is shown in Figure 2, in which the force diagram of AB rod at a certain time in the process of sliding is drawn; (2) when the speed of AB rod is V, the current and acceleration of AB rod at this time are calculated; (3) when the speed of AB rod is V, the force diagram of AB rod is obtained In the process of sliding, the maximum velocity can be reached by ab bar
(1) According to the right-hand rule, when the rod moves downward, the induced current flows from a to B. according to the left-hand rule, the direction of Ampere force on the rod AB is upward along the inclined plane, and the force on the object is as shown in the figure: gravity mg, vertical downward supporting force N, vertical inclined plane upward ampere force F, upward along the inclined plane (2) when the velocity of the rod AB is V, the force is
As shown in the figure, PQ and Mn are parallel metal guides placed in the same horizontal plane, 1m apart. A conductor bar AB is placed on the guide rail, with the mass of M = 0.2kg. The middle point of the bar is connected with the object by a rope through a pulley. The mass of the object is m = 0.3KG, the dynamic friction coefficient between the bar and the guide rail is μ = 0.5, and the magnetic induction intensity of the uniform magnetic field is b = 2T How much current should be put into the rod when the speed rises? What's the direction? (g = 10m / S2)
The sliding friction force on the conductor rod is: F = μ Mg & nbsp; & nbsp; & nbsp; and the ampere force is: F = bil. Because the object rises at a constant speed, the equilibrium condition is: bil = Mg + μ mg. Substituting the data into the solution, I = 2A & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; The direction of Ampere force is horizontal to the left. According to the left hand rule, the direction of current is from B to A. A: in order to make the object rise at a uniform speed, a current of 2A should be put into the rod, and the direction is from B to a
As shown in the figure, the fixed parallel guide rails Mn and PQ with enough length on the same horizontal plane are located in the uniform magnetic field perpendicular to the paper surface. There are two metal rods AB and Cd on the guide rail, which can slide along the guide rail without friction. The metal rod and the guide rail have a good indirect contact, and AB and CD are stationary at the beginning. Now give CD an initial velocity v0 to the right, then the following statement is correct ()
A. CD always decelerates, AB always accelerates, and is likely to catch up with CDB. CD always decelerates, AB always accelerates, but definitely can't catch up with CDC. CD first decelerates and then accelerates, AB first accelerates and then decelerates. D. CD decelerates and ab accelerates, and finally the two rods move at the same speed
According to the title, give CD a right initial velocity V0, CD rod will cut the magnetic induction line to produce induced current, this current flows through AB rod, judging by the left-hand rule, AB rod will be accelerated by the right ampere force, AB rod will also cut the magnetic induction line, so that the total electromotive force in the circuit will be reduced, the induced current will be reduced, then the two rods will be
RELATED INFORMATIONS
- 1. As shown in the figure, in a uniform magnetic field, the resistance of two parallel metal guide rails Mn and PQ is ignored, AB and CD are two metal slide rods As shown in the figure, in a uniform magnetic field, the resistance of two parallel metal guide rails Mn and PQ is not included, AB and CD are two metal slide rods, and the resistance rabf2 and UAB
- 2. As shown in figure a, the smooth parallel guide rails Mn and PQ are placed horizontally without resistance. The distance between the two guide rails is d = 10cm. The conductor rods AB and CD are placed on the guide rails and perpendicular to the guide rails. The resistance of each rod between the guide rails is r = 1.0 Ω. The two rods are tied with insulated wire with length of L = 20cm. When the whole device is in a uniform magnetic field. T = 0, the magnetic field direction is vertical and downward, and the wire is just in the ungrounded position After that, the change of magnetic induction intensity B with time t is shown in Fig. B. the influence of the induced current magnetic field is not taken into account. The whole process of the wire is not broken
- 3. As shown in the figure, the uniform electric field in a certain space is vertically downward, the uniform magnetic field is vertically inward, and a metal bar AB falls freely from high h, then () A. A end first landing B. B end first landing C. both ends simultaneously landing D. the above statements are not correct
- 4. As shown in the figure, PQ is a thin metal sheet in a uniform magnetic field, and its plane is parallel to the direction of the magnetic field. A charged particle shoots out from a certain point at a velocity V perpendicular to PQ, and the kinetic energy is e. the trajectory of the charged particle after shooting is shown in the figure. The ratio of its trajectory radius on both sides of the metal sheet is 10:9. If the resistance and electric quantity of the particle are constant in the process of passing through the plate, it can be seen as follows, Then () A. Charged particles must be positively charged. B. the velocity of charged particles decreases by 1102emc every time they pass through the metal sheet. The kinetic energy of charged particles decreases by 0.19ed every time they pass through the metal sheet. The charged particles are trapped in the metal sheet after they pass through five times
- 5. A rectangular granite is 10 meters high, 3 meters wide and 2 meters thick, with a density of 2.5 x 10 cubic kilogram per cubic meter. The maximum bearing capacity on the ground is
- 6. How many tons of tire oil can a 3 m long and 1.2 m wide rectangle hold? The density is 0.5
- 7. A granite stone with school motto has a mass of 3.18 × 104 & nbsp; kg and a density of 2.65 × 103 & nbsp; The density of brick is known to be 2.05 × 103kg / m3, and the density of cement mortar in the base brick joint is the same as that of brick (G is 10N / kg) (1) The volume of granite stone; (2) the pressure of stone on base; (3) the ratio of stone to base and base to ground
- 8. Is giraffe's heart very good? Why is its blood pressure so high Please give a scientific answer
- 9. Is a giraffe the tallest animal on land, or an elephant?
- 10. Giraffe is the tallest animal on the earth at present, so its blood pressure is also the highest among the animals. The average height of an adult giraffe can reach 4.3m, its mass can reach 2T, and the landing area of one sole of a giraffe is 250cm2? (2) If the blood density of a giraffe is 1.05 × 103kg / m3, when its heart contracts, it is necessary to press the blood to the head 3M higher than the heart, then at least how much pressure should be produced when the giraffe's heart contracts? (systolic pressure refers to the pressure produced when the heart contracts. Atmospheric pressure is not taken into account in calculation. In this problem, G is taken as 10N / kg.)
- 11. How to measure the diameter of 100cm wire
- 12. A wire with a length of 50cm is tightly wound on the pencil rod for 30 turns, and the width of 30 turns is measured with a scale, then the diameter of the wire is () mm (2) μ M
- 13. A piece of wire is wound on the pencil for 20 turns, and the width of 20 turns is 20 mm measured with a scale. The diameter of the wire is? Mm I'm in a hurry
- 14. In order to measure the diameter of the thin copper wire, a student arranged the thin copper wire tightly on the pencil for 30 turns, as shown in the figure. From the figure, we can see that the length of the coil is______ Cm, the diameter of copper wire is______ cm．
- 15. The weather is very hot. Use a balance and a scale to measure the density of a cube of uniform mass. The scale is made of a metal material which is sensitive to temperature The side length of the cube measured by the scale will be smaller than that measured by the standard room temperature
- 16. There is a mess of wire which can't be straightened out. Please use your measuring tools to measure its length quickly It is required to write out the main steps and calculation formula of the measurement method 1. Give you a scale with the minimum scale value of mm and a pallet balance Here is a round pencil, a ruler and a large cup
- 17. Measurement 1. The thickness of a piece of paper 2. The height of a building 3. The diameter of a ping-pong ball 4. The diameter of a thin wire 5. The length of a bending workpiece
- 18. The length of the wire can be obtained by knowing the cross section, density and mass of the wire That is why it is best to deduce other formulas, such as known wire length, cross section mass, metal density, etc
- 19. To measure the diameter of a thin wire, there is an 80cm wire, a pencil and a millimeter scale. Design the measurement scheme and write the results
- 20. It is known that the standard value of 1 standard atmospheric pressure is about 10 to the fifth power Pascal. Xiao Ming wants to explore the atmospheric pressure through experiments. His experimental scheme is to press the plastic hook suction cup dipped in water on a smooth horizontal table, squeeze out the air inside, and measure the reading with a spring dynamometer. This is the pressure of the atmosphere on the suction cup: then try to measure the contact area between the suction cup and the table, Then calculate the atmospheric pressure Q: Xiao Ming has chosen a spring dynamometer with a measuring range of 100 n. please explain by calculation the size of the suction cup with the bottom area at least so as to carry out the experiment smoothly? (assuming that the air in the suction cup is completely squeezed out)