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A measuring cylinder is 1m high with a bottom area of 40DM and a vertical slender tube with a length of 1.2m and a cross-sectional area of 5cm A measuring cylinder is 1m high with a bottom area of 40DM and a vertical slender pipe with a length of 1.2m and a cross-sectional area of 5cm. When the cylinder and the long pipe are filled with water, what are the water pressures on the upper and lower bottom of the cylinder and the gravity of the water in the cylinder and the pipe? (g = 10N / kg)
The pressure of water on the bottom is equal to the height of the water column multiplied by the density of water multiplied by the acceleration of gravity. The known density of water is ρ = 1kg / DM ^ 3, g = 10N / kg. The pressure on the top of the cylinder is P1 = ρ * g * H1 = 1 * 10 * 12 = 120N / DM2 = 0.012mpa. The pressure on the bottom of the cylinder is P2 = = ρ * g * h2 = 1 * 10 * (12 + 10) = 220n / DM2 = 0
Resistance and density, cross-sectional area, length formula
R = PL / s, P is not density, but a coefficient related to material. Resistance has nothing to do with density
Resistance = resistivity x Length / cross-sectional area (independent of density)
Please accept as a satisfactory answer. Ask: resistivity?
In the home circuit of physical electric power, users only use an incandescent bulb L marked with "220 V 100 W"
The actual output voltage of the incandescent bulb is only 100W. When the user uses a low-voltage incandescent bulb, the output voltage of the incandescent bulb is only 100W, How many volts should be output at both ends of the low voltage power supply
The first step is to calculate the resistance of the bulb P = u ^ 2 / R, and get the resistance of 484 Ω. The second step is to calculate the actual current p = I ^ 2R, and get the current of 0.4A. The third step is to consider the resistance value of the wire. The voltage is 220 V, the current is 0.4A, the total resistance is 550 Ω, and the wire resistance is 66 Ω
The activity of "rafting in Tianmu River" has attracted many tourists. Six logs with density of 0.45 × 103kg / m3 and volume of 0.15m3 can be used to form a raft. How many people can be on a raft?
The maximum buoyancy of raft is f = PGV row = 1.0 * 10 ^ 3 * 10 * 6 * 0.15 = 9000N
Gravity of raft: g = PVG = 0.45 × 10 & sup3; kg / M & sup3; × 6 × 0.15m & sup3; = 405n
The maximum mass of people on the raft M = g people / g = (F-G) / g = 859.5kg
Potentiometer and sliding rheostat are varistors. Their principle is to change the resistance in the access circuit by changing the length of the conductor. As shown in the figure is the structure diagram of a potentiometer, the two ends of the resistor R are respectively connected with terminals a and C, one end of the slide P is connected with terminal B through the rotating shaft o, and the other end is in good contact with the resistor R, and can slide on the resistor R around the rotating shaft o The total resistance of the resistor R is known to be 10 Ω. Connect the two terminals a and B of the potentiometer to the two ends of the power supply with a voltage of 2.4V respectively. When the slide P is in the position shown in the figure, the resistance between terminals B and C is 6 Ω
A. The voltage between terminals B and C is 2.4vb. The electric power consumed by the potentiometer is 1.44wc. The current through the resistance between a and B is 0.4ad. In the process of sliding the slide P counterclockwise to terminal C, the resistance in the access circuit becomes smaller
A. B. the power consumed by the potentiometer: P = u2rab = 2.4V × 2.4v4 Ω = 1.44w; C. The current through the potentiometer: I = Urab = 2.4v4 Ω = 0.6A; D. when the slide moves to the C end, the length of the resistance wire in the access circuit is increased, so the resistance in the access circuit becomes larger
After connecting a "220 V 40 W" light bulb and a "220 V 100 W" de light bulb in series with a 380 V circuit, what is the actual power consumption of each light bulb?
R1 = 220 ^ 2 / 40 = 1210 Ω, R2 = 220 ^ 2 / 100 = 484 Ω,
If the resistance of the lamp does not vary with temperature, then
Assuming that the lamps will not be burned out, according to the principle of voltage division in series circuit
The voltage at both ends of the "220 V 40 W" bulb u = 380 * 1210 / (1210 + 484) V is about 271 V, much larger than 220 V, so the 40 W bulb will be burned out, that is, the actual power consumption of each bulb is zero
I'm really a master. After writing so much, I didn't expect that it would burn. I've learned! Thank you, expert!
The density of log is 0.6 * 10 ^ 3 / m ^ 3. When a log of 2m long and 0.5m ^ 3 cross section is used to float on the water, how many kg can it carry at most?
100 points reward for answering at 10:00!
Suppose that the maximum capacity is XKG. According to the law of buoyancy, it is obtained that: the buoyancy of log = the weight of log + X; that is: 2 * 0.5 * 10 ^ 3 = 0.6 * 10 ^ 3 * 2 * 0.5-x (the density of water is taken as 10 ^ 3kg / M3). Solution: x = 400kg, that is, the maximum capacity is 400kg
400kg
Rheostat is a device that changes the resistance by changing the () connected to the resistance line in the circuit. The common ones are () and potentiometer
The name plate of a pulley rheostat is marked with "2A 20 Ω", in which 2A represents () and 20 Ω represents ()
Length sliding rheostat
The maximum allowable current is 2a and the maximum resistance is 20 Ω
1. Long short sliding rheostat
2。 Maximum allowable current and maximum resistance
The 220 V, 40 W and 220 V, 100 W bulbs are connected in series in the 220 V circuit to calculate the electric power and total electric power, and are connected in parallel in the 110 V circuit
Light bulb L1 (220 v-100w) and light bulb L2 (220 v-40w) are connected in series in the 220 V circuit. What is their respective electric power? What is their total power? When the lamp is 220 V and 100 W, the current passing through is I1 = P1 / u = 100 W / 220 V = 5 / 11a, resistance R1 = u / I1 = 220 / (5 / 11a) = 484 Ω. When the lamp is 220 V and 40 W, the current passing through is I1 = P1 / u = 100 W / 220 V = 5 / 11a
Now, if you want to prepare 50dm3 brine with a density of 1.1 × 103kg / m3, how many volumes of concentrated brine and clear water with a density of 1.2 × 103kg / m3 are needed? The formula and the solution idea are given. Thank you!
Let the volume of concentrated brine be V, and solve the equation
1.2 × 103kg / m3 times V is the mass of brine, 1.0 × 103kg / m3 times (0.05m3-v) is the total mass of water
The sum of the two is the total mass of brine, that is, 1.1 × 103kg / m3 multiplied by 0.05m3, and the solution is v = 0.025m3, that is to say, the concentrated brine and water need 25dm3 each
RELATED INFORMATIONS
- 1. The sliding resistor is connected to the circuit by changing the resistance______ To change______ &The size of the
- 2. 1. The actual power of a 220 V, 100 W light bulb is 2 when it is 110 v. two 220 V, 100 W light bulbs are connected in series in the circuit to calculate the power consumption after 1 min
- 3. When the constant voltage U is applied to both ends of a metal wire with uniform thickness, the current intensity passing through the metal wire is I, and the average rate of directional movement of free electrons in the metal wire is v. if the metal wire is elongated to make its length twice the original length, throw it to both ends and apply the constant voltage u, then The rate of directional movement of free electrons is v / 2 I know that the current becomes a quarter. Why does the rate become a half?
- 4. What is the specification of 2 * 2.5 square wire?
- 5. 1. A bulb connected to a 220 V circuit, the current through the bulb is 0.5 A, the power on time is 1 h, how much energy does it consume? How many kilowatt hours?
- 6. When the voltage at both ends of the conductor is 6V, the current passing through it is 0.6A, and the resistance is 10 ohm. If the voltage at both ends of the conductor is 3V, what is the current and resistance of the conductor
- 7. How much copper does a meter of 3 * 120 cable contain?
- 8. In the home circuit, the voltage must be 220 V, and the resistance of the electric stove is very large. According to P = u / R, the power should be very small, but why is the power of the electric stove very large? Moreover, the higher the resistance of other electrical appliances, the greater the power?
- 9. Does the total voltage and current in parallel circuit change with the resistance
- 10. Can two 2.5 wires be used together as a 6 square one? Is the current passing through the two wires the same? It's better to have theory, authoritative theory. I'm adding points
- 11. There are three electrical appliances. The voltage of electric fan, electric bulb and electric heater is 220 V, and the electric power is 60 W. when the rated voltage is fixed, the heat generated is relatively small
- 12. If the words "220V 10A (20a)" are marked on a household electric energy meter, it is allowed in the household circuit
- 13. In a family circuit, the electricity meter marked with "220 V 3A" is selected. Three "220 V 60W" lamps and a "220 V 75W" TV set are normally used. In the festival evening, we want to install a color string lamp marked with "220 V 15W" to add a festive atmosphere. How many lamps can we install at most? (in two ways)
- 14. The teaching building of a school is equipped with a "220V 10A" watt hour meter. How many "220a 400W" bulbs can be connected to the circuit at most? Sorry, I have the wrong number It should be a "220 V 40 W" light bulb
- 15. The rated voltage is 220 V and the rated power is 1210 W. the meter marked with "3000 R / kW * H" turns 50 r in 60 s. what is the actual voltage of the circuit? I'm going to hand in my homework tomorrow_
- 16. What is the meaning of the rated current and the rated maximum current on the watt hour meter It is said in the physics book of the eighth grade volume II of people's education press that an electric energy meter is marked with 10 (20) a, which means that the rated current is 10a and the rated maximum current is 20A. Since it is the rated maximum current, the power of the electrical appliances allowed to be used at the same time should be 4400w. Is there any difference in the rated current here? The maximum power used stably for a long time should be calculated as 10A or 20A
- 17. How to understand the 220 V 2.5 [10] a 50 Hz 1920r / kWh on the meter
- 18. The total electricity consumption is 3887220 V, 10 (60) a, 300 R / kWh, 50 Hz, and the power consumption of sub meters is 911220 V, 3 (6) a, 2000 R / kWh, 50 Hz My family shares a master meter with another family. The other family has a sub meter, but we don't have one. We always subtract their sub meter from the master meter, which is our electricity bill. I feel that their household electricity is more than ours, and our electricity bill is more than theirs. The master meter shares 3887 degrees. This is the meter of Power Supply Bureau, 220 V, 10 (60) a, 300 R / kWh, 50 Hz, 3 (6) a, 2000r / kWh, 50. They live at 911 degrees. I want to know how much electricity we have. Can we use the formula to express it?
- 19. The name plate of a household electric energy meter is marked with "NR / kW · H", which indicates that for every 1kW · h of electric energy consumed in the circuit, the turntable of the electric energy meter turns n. now there is only one bulb connected in the room (the bulb consumes 100J of electric energy per second), and the turntable turns 15 times in 3min. What is n
- 20. A household electric energy meter is marked with 220 v. a household electric energy meter is marked with "220 V 10A 3000 R / kWh". When all the electrical appliances are working, the electric energy meter turns 45 r in one minute. Therefore, the total power of all the electrical appliances in the household can be estimated as W? If the total current of all the electrical appliances in the household is not more than the rated current of the electric energy meter, the electrical appliances below w can be added to the circuit Why multiply by 10? Why divide by 3000?