Physics Master Advance!! Urgent!! I ll hand in my homework tomorrow!!

1 Incandescent lamps don't shine after a long time, why? After using it for a long time, the resistance of the filament becomes larger, because the filament becomes thinner, so the power becomes smaller.

2. The inner wall of the incandescent lamp will turn black after a long time, why? Because the filament temperature is high, it will evaporate and solidify when it is cold, so it will turn black.

3. After the incandescent lamp filament is burned out, it can be reused when it is lapped, but the light is brighter than the original, why?

But it is often easy to burn out at the lap joint, why? The filament resistance becomes smaller, so the power becomes larger and brighter, and the resistance at the lap joint is large, and the heat generated is more, and it is easy to break here.

4. When the electric furnace is connected to the circuit, the electric furnace wire is hot and red, but the wire connected to it is not very hot, why? The current is certain, the electric furnace wire resistance is large, the heat is more, the temperature is high, the wire resistance is small, and the heat generation is small, so it is not very hot!!

1st & 2nd Question: The filament of the bulb is made of metal tungsten. After being energized, the filament heats up, and the temperature is as high as more than 2500.

When tungsten metal sublimates at high temperatures, a part of the tungsten metal particles run out of the surface of the filament and settle on the inner wall of the bulb. After a long time, the bulb will turn black, and the sublimation of the filament surface causes the filament to become thinner, so the resistance becomes larger, according to the formula of power: p=u square r

The resistance increases, so the power decreases, so the light will be dimmed.

Third: After the filament is burned out, it will cause the filament to become shorter, so the resistance will become smaller, and the power of the bulb will increase according to the above formula. And at the lap joint, due to the small area of contact, the resistance of this place is large, and in the same series circuit, due to the equal current, according to the formula p=i squared * r, the local power of this place will be very high, resulting in another burnout.

4. According to the same current in the same series circuit, the power p=i square * r, the resistance of the wire is very small, so the power is very small, and the resistance of the electric furnace wire is large, so the power of the electric furnace wire is large, resulting in heating.

1. The image of the tree in the mirror moves at a speed of 72m s. Taking a person or a car or a mirror as a reference, the speed of the tree relative to the reference object is 72

The velocity of the image of the tree relative to the tree is 144m s. If you think of a tree as a reference, and the mirror moves 1 m, the resulting image moves 2 m relative to the original.

The truck is stationary relative to the shadow in the water and has a speed of 0The shadows in the lake moved with the truck, and the two of them were relatively still.

2. Select D. The referent is relative, it is the assumption that the person or object is not moving, if you can, imagine you in the position of the referent, and then look at another object, if it is moving it is moving, if it is stationary it is stationary.

3, the length of the meter) two second sound over 1, the third sound over 2, the fourth sound over 3.

Speed = Length Time =

4. Athletic.

3. Slow speed, fast is inconvenient to remember time, and the error is large.

5. The minimum scale is centimeters, then the data should be centimeters as the accurate value, followed by an estimated value, that is, the recorded data should be in millimeters, and the answer is c

The cylinder is first analyzed for force.

The vertical direction is supported by gravity and triangular prism inclined plane vertical component, the horizontal direction is supported by wall thrust and the horizontal component of triangular prism inclined plane support force, where the vertical direction force is determined.

The vertical component of the force supported by the triangular prism bevel = gravity.

Combined with the angle of the bevel, the law of parallelograms.

The triangular prism bevel support force and its horizontal components can be calculated.

The triangular prism is then analyzed for force.

The vertical direction is subject to gravity, and the vertical component of cylindrical pressure.

The horizontal direction is affected by the ground friction force and the horizontal component of the cylindrical pressure, wherein, the vertical component of the cylindrical pressure = the vertical component of the triangular prism inclined plane support force = the horizontal component of the cylindrical gravity cylindrical pressure = the horizontal component of the triangular prism inclined plane support force (action force, reaction force) Then it is easy to do, the simultaneous equation, compare the relationship between the horizontal component of the cylindrical pressure and the maximum friction force on the ground.

Among them, if sliding friction occurs, and the ground friction = friction coefficient (cylindrical gravity + triangular prism gravity) is not uploaded**, I can only talk to you about qualitative analysis.

But it should have been made clear.

Solution: (1) If S1 is disconnected and S2 is closed, only R1 is connected to the circuit U IR1 1A 6 6V

2) When S1 and S2 are closed, R1 and R2 are connected in parallel, R3 and the voltmeter are short-circuited, and the voltage representation number is zero.

1/r=1/r1+1/r2

R4 3) If S1 and S2 are disconnected, R1 is connected in series with R3.

When the current is current, the resistance of R3 is the smallest.

r total u i 6v 10

r3 small r total r1 10 6 4

When the voltage is 3V, R3 has the largest resistance.

u1＝u－u3＝6v－3v＝3v

iSmall U1 R1 3V 6

R3 is large U3 i small 3V

The R3 resistance varies between 4 6.

1, S1 is disconnected, R2 is broken, R3 voltage is shorted by S2, and the power supply voltage is 6*1=6V

2, s1 s2 closed.

R1 R2 is connected in parallel, R3 and the voltage mark are S2 short circuit resistance is 1 R=1 R1+1 R2 R=4 3, S1 S2 is disconnected in series, that is, R1 and R3 are connected in series, and the voltmeter is measuring the voltage of R3 due to the maximum 3V of the voltmeter

And know that the voltage U is 6V

Yes: u (r1+r3)=3 r3

And when the voltmeter is 0, it is r3=0

The r3 range can be found.

You can draw your own drawings in the calculations.

That's easy to understand.

Hope that helps.

(1) There is only R1 in the circuit

e=u1=i*r1=1a*6ohm=6v

2) The total resistance is R1 and R2 in parallel (the sliding rheostat is short-circuited), R=R1R2 (R1+R2)=4 ohms.

3) U3<=3V, so R3<=6 ohms.

i<=, so r3>=4 ohms.

Therefore, 3< = r3< = 6 ohms.

1. 6v

2.(Excuse me, where is the rheostat slide position?) )

2. （1） i=4\20=

2) U=10 ohms*

3) Exist When the voltage division of the ammeter is not considered, when p slides to the leftmost end, there , vi is equal to v, and the angle is the same when the range is the same resistance is 20 ohms When p does not slide to the endpoint, it does not exist, v is the total voltage of r1 and r2, and it is also the power supply voltage, v1 is the voltage of r1 Since the two voltmeters are not equal at this time, to make their angles equal, the range should be different, v is 0-15v, v1 is 0-3v At this time, v is 6v, v1 should be 3*(6 15)=

i.e. r2=(ohm 80》20 so it does not exist.

You're a junior high school student, the question is a review question, I'll give you an idea: think about what formulas are involved, so that you won't be unable to touch the door (power is at the end of the second year of junior high school, and electricity is at the end of the third year of junior high school).

The slide is the same at v and v1 at the far left, where the total resistance is equal to 20

The total power of all the bulbs in the "Bird's Nest" is about 1000 kW.

1.The indication of the ammeter.

2.The indication of voltmeter v. 6v

3.In the case that the range of the meter can be changed, is it possible to change the position of the slide p so that the two voltmeter pointers deviate exactly the same angle from the zero scale? If this is not possible, please state the reasons; If possible, calculate the total resistance in the circuit.

Probably, the total resistance is 20 ohms. (r2=0 Ω satisfied).

This one is too complicated. I won't.

2.The indication of the ammeter.

The voltmeter shows 6V

3.No.

Because: V is the external voltage of the power supply, V1 is the R1 voltage, the ammeter and the sliding rheostat will divide the voltage. (Of course, it is possible to ignore the partial pressure).

1.The indication of the ammeter 2 of the voltmeter V is 6v 3 possible, and the total resistance is 20

1000kw

2.The indication of the ammeter.

The voltmeter shows 6V

You're going to have to study hard ... It's all basic.

Because w=pt and w=1000kwh, t=1h, p=1000kw) 2.The indication of the ammeter.

i=u/r=4v/20ω=

The number of the voltmeter V is the sum of the voltage of the V1 indicator and the rheostat (in series) U = 4V + the sliding vane is in the middle, and only half of the resistance is connected to the circuit) 3.No.

Because: the number of voltmeter V is also the power supply voltage, which is constant, always 6V, and now the range of the two voltmeters is 0 15V. In order for the two voltmeter pointers to point to the same position, the range of the voltmeter v1 must be changed to the range of 0 3v, and the indication is that the position of the two pointers is the same.

Let the effective resistance of the sliding rheostat be r at this time. Then the two resistors are connected in series with equal currents. Yes

then r = 80 , which exceeds the maximum resistance of the sliding rheostat.

1. p=1000 kWh, t=1 hour w=p*t=1000*1000*1000 watts.

2、r1=20ω r2=10ω u1=4v

i1=u1 r1=4 20= i=i1=i2 u2=i2*r2= u=u1+u2=6v because of the series

If the two voltmeter pointers deviate from the zero scale at exactly the same angle, it means that the two meters use different ranges, assuming that the pointers of the two voltmeters are deviated by a grid, then the reading of voltmeter 1 u1 = voltmeter reading u= then u2=

i1=i2=u1/r1= r2=u2/i2= ω

Answer: 1. The indication of the ammeter and the indication of the voltmeter V are not possible, because if it deviates from the same angle, the resistance value of the sliding rheostat is calculated to be 80, and the sliding rheostat R2 is marked with the words "20, 2A", and the resistance value does not exceed 20, so there is no such possibility.

Question 1: 1000kw

Question 2: The resistor R1 and the sliding rheostat are connected in series, the voltmeter V1 measures the voltage at both ends of R1, the V measures the power supply voltage, and the ammeter measures the current in the dry circuit.

1) from Ohm's law i=u r; i=u1 r1=4v 20 euros=

2) Voltmeter V measures the power supply voltage, that is, the total voltage, U=U1+U2=4V+10 ohms.

3) There may be two cases where the angle of the two tables deviates from the zero scale exactly the same: one is u1=u, and u=5u1 (because v1 uses 0 3v and v uses 0 15v, and the relationship between these two readings is 5 times).

When U1=U, that is, the sliding rheostat slides to the far left, the sliding rheostat is short-circuited, and R1 is connected in series with the power supply, then the total resistance at this time is 10 ohms.

When U=5U1, that is, U1=, current I1=Euro=, the voltage at both ends of R2 is: U2=U-U1=, then R2=U2 I1=Ohm, and the maximum resistance of the sliding rheostat is 20 ohms, then this situation does not exist.

I get it.

Figure A is the voltmeter v is 15V range, Figure B is the voltmeter v1 is 3V range, so U A = 6V, U B =

So u=uA-uB=

So i=u r=

In the diagram v=, v1=, then the voltage on r is vr = current i = vr r = amperes).

It is necessary to determine the range.

From the circuit diagram, you know that you are greater than U1

Therefore u=6v,u1=

U Resistance = So i = u Resistance r=

So the voltmeter V is connected.

The acceleration of the electron is perpendicular to the metal plate, and the area of the electron hitting the B plate is required, and only the maximum distance xmax that the electron can move in the direction of the parallel metal plate is required, and then the maximum area can be calculated with s= xmax 2.

First of all, it can be judged that when the initial velocity is parallel to the metal plate, x is the maximum.

This is the base dismantling because, in the parallel metal plate direction x=v flat *t

If the initial velocity has a component in the direction of the perpendicular sheet metal, then the relative initial velocity of t will decrease when it has a component in the direction of the perpendicular sheet metal, and the v-level will also decrease. Therefore, it is known that x is the maximum when the initial velocity is parallel to the metal plate.

Then you can ask for xmax in the dust front.

The acceleration of an electron a=ue md

In the direction of the vertical pie wheel sheet metal, at 2 2 = d

t = root number (2md ue).

then xmax=vt=v times the root number (2md ue).

Then s=xmax2=2 mdv2u

Position where the field strength is zero:

1. "The same charge is in between" and "different charge is on both sides".

2. Stay close to the charge with "less charge".

Let the coordinate at the zero field strength be x (to the right of q2).

The field strength of q1 at x is e1=k*q1 x 2

The field strength of q2 at x is e2=k*q2 (x-6) 2 is obtained by e1=e2:

k*q1 x 2=k*q2 (x-6) 2, substituting q1 and q2 to get the solution:

x=12cm

1) At x>6cm.

e1-e2=k*q1 x 2-k*q2 laughing stupid (x-6) 2k(10 -8).

In the above formula, when 》0, e1-e2>0

And x>6, so take x>12cm

2) At all points on the x-axis of 012cm", the combined field strength is made in the direction of "to the right" (along the positive direction of the x-axis).