Exercise 1 This is a typical problem in judging sliding rheostat. At first, many students thought that the resistance of the sliding rheostat connected to the circuit would increase. But careful observation, using the principle of circuit simplification, regards the ammeter as a short circuit and the voltmeter as an open circuit. We can find that the current does not pass through the slider, and the sliding rheostat does not change the resistance, so the current in the circuit will not change. However, because the slider moves to the right, the resistance range measured by voltmeter becomes smaller.
In Exercise 2, the lamp is connected in series with the sliding rheostat, the ammeter measures the current in the series circuit, and the voltmeter measures the voltage across the lamp. When moving to the right, the resistance in the circuit increases, the current decreases, and the filament resistance remains unchanged. As the current decreases, the electric lamp dims, so when P moves to the right, the ammeter reading decreases, the voltmeter reading decreases, and the electric lamp dims. C is correct.
Exercise 3 As can be seen from the figure, the constant resistance is connected in series with the rheostat, and the voltmeter is connected in parallel with the rheostat, so the reading ratio of the voltmeter and ammeter is the value of the rheostat connected to the circuit. When the slider P moves to the bottom, the resistance of the rheostat connected to the circuit increases, the current in the circuit decreases, the reading of the ammeter decreases, and the resistance remains unchanged. It is known that the voltage at both ends becomes smaller, and the reading ratio between voltmeter and two meters becomes larger.
Lian 4B
Practice 5 When the voltmeter reads zero, the slider should be at the far left. At this time, the current in the circuit is the largest, and the ammeter reads:, and the power supply voltage.
When the voltmeter reads 4V, the slider P should be at the far right.
At this point, the ammeter reads, power supply voltage, and then
In other words, the solution.
It is a common method to measure voltage step by step with voltmeter. Both lights are out, and the ammeter has no obvious deflection. These two phenomena reflect that the current in the circuit is too small, but it also reflects that the ammeter is in the path, and it is impossible that the L2 filament is burnt out or the lamp cap is not connected properly. Again, exclude a and d. Because the voltmeter is obviously deflected, it means that the voltmeter is on the path. When there is an open circuit in the circuit of aL 1b, the voltmeter is connected in series with the circuit of ammeter L2 and power supply, resulting in a large total resistance of the circuit (generally, the internal resistance of voltmeter is close to104Ω), so that the pointer of ammeter hardly moves and L2 lamp does not light up. Therefore, the fault occurred in the L 1 branch. Answer: B.
Exercise 7( 1) The lamp L is short-circuited or the resistance is open;
(2) the lamp L is short-circuited; Resistance is open.
(3) no. In the case of short circuit of lamp L and open circuit of resistance, voltmeter V has no indication.
Practice 8C
Exercise 9 bulb open circuit
Exercise 10B
Exercise 1 1B
Lian 12D
Exercise13 (1) c (2) d.
Practice 14EDCABHFG
Practice the protection at the maximum resistance of 15( 1) and change the current of the circuit (2) The resistance of the small bulb changes with the temperature.
Safe use of electricity, electricity and magnetism.
The even 1 switch should be connected in series between the electric lamp and the live wire. When the switch is turned off, the light is no longer connected to the live wire, which is safer. Don't put the switch between the light and the zero line, so the light of the switch is still connected to the live wire, and people are easy to get an electric shock once they touch the lamp holder; You can't connect the switch directly between the live wire and the neutral wire, so closing the switch will form a short circuit, which is very dangerous. The connection mode of three hole socket is "Left Zero, Right Fire, Shangdi". In this way, when the three-pin plug of the electrical appliance is inserted into the three hole socket, besides connecting the electrical appliance to the household circuit, the shell of the electrical appliance is also grounded. Even if the casing leaks electricity, the current will flow away from the grounding conductor, and the human body will not get an electric shock when touching the electrical casing.
Practice 2 three hole socket's upper hole ground wire, the left hole is connected to the zero line, and the right hole is connected to the live wire, so just connect the upper hole to E. Because this jack ground wire, even if the washing machine shell is charged, it can be grounded smoothly to ensure safety. The connection is shown in the figure. If the wire of this jack is broken, once the insulation between the electrical shell and the live wire is destroyed, it is easy to cause an electric shock accident.
Practice 3 Because the fire situation is not very serious at this time, Xiaohua should quickly turn off the main switch in his home (there is no need to turn off the switch of the whole building, which will affect other users. ) Then quickly put out the fire with dry powder fire extinguisher or sand, depending on whether you need to call 1 19 for help.
Practice for 4 days
Practice 5C
There are two reasons for fuse blowing: short circuit or excessive total power of electrical appliances. There is no blown fuse when the plug is inserted into the socket, so it is certain that there is no short circuit between the plug and the socket; If the switch is short-circuited, only the bulb will glow when the plug is inserted into the socket, and the fuse will not blow, so the fault may be caused by the short circuit of the lamp holder. Answer: D.
As can be seen from the picture, there may always be useful appliances working between the live wire and the neutral wire, because there is no power failure. For A, although one hand touches the zero line, standing on the ground can make the current form a loop, so it is possible to get an electric shock at this time. Therefore, d is correct.
The rated current is the current to ensure the normal operation of the circuit, and the fuse current is slightly larger than the rated current, so the answer is C.
Practice 9A, b, c
Exercise 10( 1) Maximum power allowed to access the circuit.
Connected power supply
The power of the appliance that allows reconnection is
The number of lights that can still be connected
So the number of lights that can be connected is 18.
(2), so you can connect a "220V 500W" electric furnace.
After the bar magnet 1 1 is broken into two sections, each section has a magnetic pole and is magnetic, which becomes two small bar magnets. Therefore, D.
The iron ball of Lian 12 is hung under the spring dynamometer and will be magnetized when it is close to the magnet. The magnetism near the magnet pole is strong, but the magnetism in the middle of the bar magnet is weak. The iron ball is a soft magnet, and its magnetism is easy to disappear, so the mutual attraction at the magnetic poles is large, and the spring dynamometer shows a large number; In the middle of the bar magnet, because there is no magnetism and almost no attraction, the pointer of the spring dynamometer becomes smaller.
Exercise 13 After two nails are attracted, they are magnetized by a magnet. One end of them attracted by the magnet is a magnetic pole with the same name, which is different from this end of the magnet. According to the nature of the magnet, the other ends of the two nails are also magnetic poles with the same name. Magnetic poles with the same name are mutually exclusive and should be separated. Answer: B.
There are two ways to practice 14. 1. According to the interaction between magnetic poles, it is judged that the right end of Figure A is the S pole and the left end is the N pole. The right end of Figure B is the N pole and the left end is the S pole. Then, according to the magnetic induction lines around the magnet, the directions of magnetic induction lines of A and B can be marked. Another way of thinking is that according to a small magnetic needle placed in a magnetic field, when it is at rest, the direction pointed by the N pole is the direction of the magnetic field at that point, that is, the direction of the magnetic induction line at that point. The magnetic induction lines around the magnet all come out from the north pole of the magnet and return to the magnet.
Exercise 15 First, according to the magnetic pole of the solenoid, determine the current direction in the solenoid with ampere rule (you can draw it gently at the top of the figure), and then draw the current end flowing from the positive pole of the power supply to see if the wire meets the requirements, whether it is from front to back or from back to front. Draw the first turn, then draw several turns in turn, so that the current flows back to the negative pole of the power supply, and finally check with the ampere rule, as shown in Figure B. 。
power
Exercise 15( 1) When heating at high temperature, the current through R2 is I2 = I2 = U/R2 = 220v/80ω= 2.5a
(2) At low temperature, only R 1 is connected to the circuit, and its power is 40W, so R1= U2/p = (220v) 2/40w =1210ω.
(3) When cooking at high temperature, R 1 and R2 are connected in parallel, and the power of R2 is P2=UI2=220V×2.5A=550W.
Then the total power of the circuit is P=P 1+P2=40W+550W=590W.
Therefore, the consumed electric energy W=Pt=0.59kW×0.5h=0.295kWh.
Pressure, pressure
Practice 1 as shown in the figure.
Lian 2A
After three years of training, people's pressure on Apple is certain. Cutting the apple with a knife reduces the stress area and increases the pressure on the apple. In B and D, the steel wire is thin and the needle is sharp. The purpose is to reduce the stress area and increase the pressure under the condition of constant pressure. The answer to the third grade physics exercise book in C puts the rail on the sleeper instead of directly touching the ground, so the pressure on the ground is constant, but the stress area between the rail and the ground becomes larger, so the pressure on the ground decreases at this time.
Exercise 4C
Practice 5C
Practice 6 The same brick, no matter how it is placed, is still on the horizontal ground. The gravity of the brick and the supporting force of the ground on the brick are a pair of balanced forces, so the supporting force is equal to gravity, and the action of the forces is mutual, so the pressure of the brick on the ground is equal to the gravity of the brick. The gravity of the same object is constant, that is, the pressure of brick on the ground is equal in three cases. Options A, B and C in the question are all wrong. Only option d is correct. In order to further prove the correctness of option D, it is necessary to analyze the pressure of brick on the ground. According to the calculation formula of pressure, the smaller the stress area, the greater the pressure, so the pressure on the ground is the greatest when the brick is placed vertically. Answer: D.
Practice 7B
Lian 8
Practice 9C
Exercise 10 to determine the weight of criminals is to ask people about the pressure on footprints. When the pressure is known, the key is how to find the footprint area of irregular patterns. According to the knowledge of density, we can know the mass, thickness and density of wax shoe mold, and then find the shoe mold area.
The volume of wax shoe mold is obtained by density formula ρ = m/v.
V wax = m wax/ρ wax = = 0.75× 10-3m3.
Then the shoe mold area s 1 = v wax /d = 2.5× 10-2m2.
When a criminal stands, his feet touch the ground.
S = 2s 1 = 2.5× 10-2 m2×2 = 5× 10-2 m2
The pressure is obtained by the pressure formula p = f/s.
f = pS = 1.5× 104 pa×5× 10-2 m2 = 750n
The pressure on the criminal beach is equal to the weight f = g, so the weight of the criminal is 750 n.
Exercise 1 1( 1),
(2)
Exercise121550 pa; 2× 104Pa
Exercise1320 n; 4× 108Pa
Exercise1435 n; 5000Pa2.3× 104Pa
Exercise15150n; 15000Pa
atmospheric pressure
Exercise 1C
In order to prevent cans from deteriorating, manufacturers usually vacuum tin-capped cans in glass bottles. Because the pressure inside the bottle is less than the atmospheric pressure outside the bottle, it is difficult to open the tin can on the physics exercise book for grade three. After prying open the bottle with a screwdriver, it is easy to open it. This is mainly because a small amount of air enters after the bottle is pried open with a screwdriver, which can reduce the pressure difference between the gas inside and outside the bottle. Answer: C.
Exercise 3 when the temperature is constant, the smaller the volume of a certain mass of gas, the greater the pressure; The larger the volume, the smaller the pressure. When we inhale, the chest cavity expands and the alveoli in the chest cavity expand, so the volume of the lungs increases and the air pressure in the lungs decreases, forming a negative pressure area, which is smaller than the atmospheric pressure in vitro. Atmospheric pressure forces fresh air into the lungs through the nasal cavity and trachea, while when we exhale, it is the opposite. Answer a.
A thermos bottle with a certain amount of water is sealed with a certain amount of gas. After a period of time, the temperature in the bottle is lower than when hot water was just poured, and the air pressure becomes smaller. The air pressure in the bottle is less than the outside air pressure, so the cork is pressed tightly, and the cork is not easy to pull out. Answer: B.
Exercise 5 "Pressing" the spring leaf means that part of the air in the rubber tube is squeezed out. After being released, it means that the volume of the remaining air in the tube becomes larger and the air pressure becomes smaller, which is less than the external atmospheric pressure. Understand these, and the problem will be solved. Answer: When the spring piece is pressed, the air in the rubber tube is squeezed out. After being released, the volume of the remaining air in the tube becomes larger and the air pressure becomes smaller. The air pressure on the surface of the ink is greater than that in the tube, so the ink is pressed into the rubber tube.
The standard atmospheric pressure of 6 1 is constant and does not change with the length of the glass tube. If the top of the test tube is opened, the upper surface of mercury in the tube is also affected by atmospheric pressure, which is equal to the atmospheric pressure on the surface of mercury in the mercury tank outside the tube, and mercury falls due to gravity.
Practice 7B
Lian 8
Exercise 9 To solve this problem, you can first estimate the area of the upper surface of the palm as 150cm2, and then,
So ... Answer: C.
In exercise 10, we only need to calculate the stress areas of the two hemispheres first, and then calculate the pressure of atmospheric pressure on the hemispheres, and then we can calculate the number of horses needed. Then, the stress area of the hemisphere is m2, and the atmospheric pressure is Pa.
So the pressure on the hemisphere
The number of horses used to open the hemisphere is: (horses), and the answer is: 16 horses.
The joint 1 1 pump pumps water by atmospheric pressure. We can calculate the height at which the water in the pump rises under the action of atmospheric pressure. Suppose the outside world is a standard atmospheric pressure, and the maximum height that the pump can lift water is H. At this time, there is no air under the piston of the pump, and the atmospheric pressure should be equal to the pressure generated by the water column.
P0 = ρ mercury h mercury g = ρ water h water g,
That is to say, under the action of atmospheric pressure, the water in the pump can only rise by about 10 meter, so it is impossible to pump water 20 meters deep at a time.
Exercise 12( 1) becomes larger; (2) small; (3) small; (4) getting bigger.
Exercise 13 answer: C.
Practice 14 water to boil again; The air pressure on the liquid surface decreases and the boiling point of water decreases.
Exercise 15 The greater the air velocity on the surface of an object, the smaller the pressure on the object, and the smaller the air velocity on the surface of the object, the greater the pressure on the object. The upper part of the wing is curved and the lower part is straight. At the same time, the distance of the upper airflow is longer than that of the lower airflow, so the upper airflow has high speed, low pressure and large pressure difference, and the plane is lifted up.
Lian 16D
Exercise 17. Choose D correctly.
Due to atmospheric pressure, water can rise from the thin tube A. The air pressure above the water level in the cup is constant. Only when the air pressure above the water surface in the A pipe decreases will the water rise along the suction pipe. When the air flows out of the small hole of the B tube quickly, the air near the small hole has a high velocity and a low pressure, while the air pressure above the liquid level in the container remains unchanged. Under the action of air pressure, the liquid will rise along the pipe A and flow out of the pipe. This principle has many applications in life, such as various sprayers, perfume atomizing bottles, and devices for mixing gas and air in domestic gas stoves.
buoyancy
Exercise 1 As long as an object is immersed in liquid or gas, it will be subjected to buoyancy in the vertical direction. Therefore, people who swim by the sea will be buoyed by the seawater. The answer to the third grade physics exercise book is: the hydrogen balloon floating in the air is buoyed by the air; When a stone is thrown into the water, it will also be buoyed by the water during the sinking process; The wooden blocks floating in the water are also affected by buoyancy. So right, c is wrong. The stone sinks because gravity is greater than buoyancy. C is correct.
Exercise 2 is shown in the picture.
Exercise 3C
Although the shapes of four objects A, B, C and D are different, the types of objects are different, some are solid, some are hollow, and the depth of immersion in water is different, but these different situations have nothing to do with buoyancy, so they need not be considered. Buoyancy is only related to the density of the liquid and the volume of the displaced liquid, because they are immersed in the same liquid, so buoyancy depends on it, but the D object is closely combined with the bottom of the container.
Draw a picture according to the meaning of the question. Aluminum block is a regular object, and its depth in water is known. Buoyancy can be obtained by using the pressure difference between the upper and lower surfaces of aluminum blocks. In addition, the volume of aluminum block can be obtained, because it is submerged in water and the liquid is water, so the density is known, so it is obtained by Archimedes principle.
Exercise 9 When the plastic ball floats on the water, according to the ups and downs of the object, we can know that buoyancy is equal to gravity. If ethanol is slowly added into the beaker and stirred at the same time, the density of the liquid will be smaller than that of water and larger than that of ethanol, so it may be greater than, equal to or less than. If or, the ball will float or suspend, and the buoyancy is still equal to gravity, so the buoyancy remains unchanged, then A, C and D are wrong; If the plastic ball will sink to the bottom of the cup and the buoyancy will be less than gravity, then B is correct.
Exercise 10 The buoyancy of the wooden ball and the iron ball cannot be directly compared by Archimedes principle, because although the volume of the wooden ball is larger than that of the iron ball, the volume of the underwater part of the wooden ball cannot be directly compared with that of the iron ball because it floats on the water. This problem can take the same mass (gravity) as a comparison bridge, and draw a conclusion by using the relationship between buoyancy and gravity in the case of ups and downs. The buoyancy of wooden balls in water is equal to gravity, while the buoyancy of iron balls is less than gravity.
Exercise 1 1 Frog floats in water, and its buoyancy is equal to its own gravity, but it sinks in another liquid, and its buoyancy is less than its own gravity, so its buoyancy in water is greater than that in another liquid.
Exercise 12 If an object sinks into the water and the volume of boiling water is, it can be calculated that the buoyancy of the object when it sinks into the water is, that is, the maximum buoyancy of the object after entering the water is less than its gravity, so it can be judged that the object will sink to the bottom after entering the water, and the buoyancy at this time is 4N. The answer is correct.
Exercise 13 according to Archimedes' principle, buoyancy is equal to the weight of the displaced liquid, so the buoyancy of the wood block is, and the weight of the wood block is also known through force analysis; The volume of the block is, so the density of the block is; The magnitude of buoyancy is the pressure difference between the upper and lower surfaces of the object, so the pressure on the lower surface of the object is 0 and 0. The answers are a, b and C.