Physics questions for the fourth year of junior high school, and practice questions for the fourth y

Q suction = cm (t-t0).

q Suction represents the heat absorbed, c represents the specific heat capacity, m represents the mass, t represents the final temperature, and t0 represents the initial temperature.

q amp = cm (t0-t).

Q represents the heat discharged, c represents the specific heat capacity, m represents the mass, t represents the final temperature, and t0 represents the initial temperature.

1. Heat absorbed by water: Q absorption = cm (T-T0) 4200x20x (55-25) = 2520000J

The internal energy of water increased by 2,520,000 J

2. Q suction = cm (t-t0).

Fourth power t=110°C

But at a standard atmospheric pressure, the water temperature can only reach a maximum of 100 °C, at which point it is already boiling, and the temperature will not rise again.

So the temperature of the water rises to 100°C

Please click on the lottery and ask if you are not sure.

w=cm（t1-t2）=

In the same way: ,,20+90=110, because whose boiling point is 100, so the water temperature rises to 100

First, calculate their rated currents, i=p u, to get 1 3 A and 2 3 A, respectively.

Since it is in series, in order to achieve the meaning of the title, then the current should be 1 3A, so that the first one emits normally, and the second one is darker.

Then calculate their resistances, r=u2p, and get 18 ohms and 18 ohms, respectively.

So the total resistance is 36 ohms, and the voltage at both ends is u=i*r, resulting in a voltage of 12 V

In the same way, the total power p=u*i, p= 4 w can be obtained. Finished.

One glows normally and the other is dim, indicating that the current is giving the lower power one.

The two bulbs are rated at 1 3 amps and 2 3 amps, respectively, with a current of 1 3 amps.

Both bulb resistors are 18 ohms.

The voltage is 6V each.

The voltage at both ends is 6+6=12V, and the total power is 4W for their respective current times and voltages.

Upstairs is so fast.

C If it's right, how can B be right, the water vapor is transparent and can't be seen, all you see is water mist, but what comes out of the kettle is the water mist condensed by water vapor meets dust, and the popsicle is the water mist condensed by water vapor in liquefied air.

Option C Water vapor is invisible and will be permeated around the popsicle taken out of the refrigerator"White gas"It is the original hot water vapor in the air that liquefies into small water droplets; And when the water boils, the spout will squirt out"White gas";It is the water vapor that it spews out that liquefies into small droplets when it is cold, so you can't see the white gas near the nozzle";Because the water vapor is not yet fully liquefied.

1. Water vapor is transparent and cannot be seen, and all you see is water mist.

The popsicle is surrounded by "white gas", which is the small water droplets formed by the original water vapor liquefaction in the air, and the spout will spray out the "white gas" after the water boils, which is the small water droplets formed by the liquefaction of the water vapor that is sprayed, not the original water vapor liquefaction in the air;

b is the condensation of hot air, which is somewhat similar to the formation of rain, not water vapor.

Q water suction = C water M water (t-t0) = (kg· ) x1000kgx（50 -25 ）=

If the coal saving m kilogram is saved, then according to q coal = m coal q coal and q coal x 10% = q water absorption.

m = kilogram can be found.

Since there is no picture, you can only tell the following judgment method, and you can judge it yourself.

7. B or C, the one with a larger resistance.

5. B or C, the lamp connected in parallel with the voltmeter is open.

4. C (when connecting C and D, both lights are on, indicating that both lights are fine, excluding A, B, D) 8, D (resistance is a property of conductors, and has nothing to do with voltage and current, you go to the electronic city to buy resistors, those resistors are not connected to the circuit, but they are marked with how many resistances).

9. A (in series, the current is equal, and the voltage is proportional to the resistance.)B (The resistance of 6V and 12V is respectively, and the total resistance in series in 15V is 60, then the current is, which is the same as the rated current of 12V. ）

The resistance is small, and the filament is thick); 5:3 (the power ratio in series is the opposite of that in parallel).

1. When the sliding piece p is located at the midpoint c, half of the constant resistance r0 and the resistance r r of the sliding rheostat are connected in series. When the slider p is at point B, the constant resistance R0 and the resistance R of the sliding rheostat are connected in series.

Since the series currents are equal everywhere, the total voltage is equal to the sum of the voltages of the parts. The voltage of the power supply is kept constant.

So 2 (1 2)r=(u-2) r0; 3/r=（u-3）/r0；The solution is r0=r.

2. When the slider p is located at point B, the power of the fixed resistance R0.

Substituting r0=r into the equation yields r0=30( ).

3. Substituting r=r0=30 into either of the previous two equations gives u=6v.

When the sliding rheostat slider p is located at the midpoint C, the dry current is I1, and the maximum resistance of the sliding rheostat is R, according to Ohm's law.

i1(r0+r 2)=e (1) e is the supply voltage.

i1r/2=2 （2）

When the slide p is located at point B, let the trunk current be i2, according to Ohm's law.

i2(r0+r)=e (3) e is the supply voltage.

i2r=3 （4）

i2^2r= （5）

i2 = r = 30 from (4) (5) and i1 = 2 15a from (2) (4).

Substituting equation (1) and (2) yields r0=30 e=6v

In the circuit shown in the figure, the power supply voltage remains unchanged, the switch is closed, and when the sliding rheostat slider p is located at the midpoint C, the voltage representation number is 2V; When the slider p is located at point B, the voltage number is 3V, and the power of the fixed resistor R0 is , then the power supply voltage is.

6V, the constant value of the resistance R0 is 30

The resistance of the rheostat is set to 2r. From the first condition, r (r0+r)*u=2;From the second condition, 2r (r0+2r)*u=3;From the third condition, u (r0+2r)=i; （i^2）*（r0）=；r0=2r for the solution of the three equations; u=6;r=15;r0=30。I don't know if there is a solution to the equation, and I hope to adopt it.

1.The electric resistance of the magnet current is strong in the number of turns of the solenoid.

2.It can be talked about to control the strength and weakness of magnetic Chang's destructiveness.

The iron core has or is not an electromagnet.

1) Without R1, the indicator light is not on.

2) Before boiling, p=U2 r3=2202

After boiling, p=U2 (R2+R3)=500W3) must choose 6V 10mA light-emitting diode, because the indicator light is used to indicate, the smaller the power, the more energy-saving, the bulb power, and the light-emitting diode only,