Finding a physical problem, solving the process, solving a physical problem

1) When the motor does not rotate, the motor is equivalent to pure resistance, and the coil resistance of the motor can be obtained according to Ohm's law;

2) According to the formula of power p=ui, the power of the motor can be obtained, and then the efficiency of the motor can be obtained;

3) When the rotor of the motor is stuck, the motor at this time is equivalent to a pure resistance, and the power generated can be calculated according to p=u r.

If you need a detailed process, please ask.

Solution: (1) when the motor does not rotate, let the coil resistance be r, then r=u1 i1 = ,2) the power consumed by the motor when it is working normally is p = u2 i2 = w = w, the power of the coil resistance loss when the motor is working normally is p loss = i2 r = w = w, the output power of the motor is p out = p - p loss = w = w, the efficiency of the motor is = p out p 100% = when the motor rotor is stuck, all the electric work is converted into electric heat p heat = u2 r =

1) The resistance of the electric coil r r=u i=

2) When the motor is working normally, the output power p=u*i=

and the efficiency of the vacuum cleaner =p useful p total = (ui-i 2r) ui = (3) If the rotor is suddenly stuck when the motor is working normally, the heating power of the motor at this time.

pq=i^2*r=1^2*

Hints for you. When it does not rotate, it is the coil resistance.

It can be used to answer two or three questions.

The second question can be answered directly.

First find the resistance r=u i=ohm, the heat power p heat = i2 squared multiplied by r = in normal operation, and the total power p = ui = 2w, so the output power of the motor p = p total - p heat =, so mechanical efficiency = p output p total =. When the motor is stuck, the passing current i=u r=4a, and the thermal power at this time is pheat=ptotal=ui=8w

Here's how it works, I don't know if it's right or not.

According to the formula, elasticity f = stiffness coefficient k * this variable x

Then, let the original length of the spring be l

f1=k*（l-l1）

f2=k*(l2-l)

Add the two formulas. f1+f2=k*(l1+l2)

then, k (f1+f2) (l1+l2).

f1=kl1

f2=kl2

Add the two formulas. So the stiffness coefficient k f1+f2 (l1+l2).

After the two balls absorb equal heat, there is q c a*m a* (t a t0), t0 is the initial temperature q c b * m b * (t b t0).

From the latter condition "Ball A transfers heat to ball B", it can be seen that the final temperature t A t B has C A*m C B * M B.

What can happen is:

1. C A C B, M A M B.

2. C A C B, M A M B.

3. C A C B, M A M B.

4. C A C B, M A M B.

5. C A C B, M A M B.

q=m*c*δt, the heat absorption is the same, and A can transfer heat to B, then the temperature rise of A is greater than the temperature rise of B, so M A*C A

The condition for determining is that "after the two balls absorb equal heat, ball A transfers heat to ball B", so there will be many cases of this problem, and the column formula is.

m A * C A * T A = M B * C B * T B (where T A, T B represent the temperature rise) and T A" t B, so M A * C A < M B * C B then discuss it However, it is estimated that your topic is only required to write this formula, and it is impossible to ask you to find the mass ratio and the specific heat capacity ratio separately.

That's right, the answer is wrong.

2. Let the area be s, and obtain s=225cm 2p=f s by =m v and v=, and get p=26000 pa.

It is 3cm when it is trapped in the mud, so the pressure p is the same.

f=ps, f=585n, so mass=

During the infusion, the liquid should be dripped at a uniform rate, which is essentially required to maintain the pressure at the mouth of the infusion bottle during the continuous decrease of the liquid medicine in the bottle. In Figure A, the gas above the liquid medicine communicates with the atmosphere, and the pressure in the liquid at the mouth of the bottle is equal to the sum of the atmospheric pressure and the pressure generated by the liquid in the bottle. In Figure C, when the sum of the pressure generated by the liquid medicine and the pressure generated by the gas in the bottle is less than the external atmospheric pressure, the pressure in the liquid will enter the bottle, so that the pressure in the liquid at the mouth of the bottle is always basically equal to the atmospheric pressure, so that the liquid can drop at a uniform speed; In Figure B, when the liquid level in the bottle drops, the gas outside the bottle can enter the bottle through the air inlet pipe under the action of atmospheric pressure, and the liquid medicine can drop at a uniform speed for a period of time, but once the liquid medicine drops below the mouth of the air inlet pipe in the bottle, it is the same as Figure A, and the liquid medicine cannot be kept to drip at a uniform speed throughout the infusion process; In Figure D, there is only the outlet of the liquid medicine, as the liquid medicine enters the body, the pressure of the gas in the bottle decreases, and the pressure generated by the liquid itself also decreases, so that the pressure in the liquid at the mouth of the bottle decreases quickly, and finally the liquid pressure at the mouth of the bottle is small and the liquid cannot be injected into the body. Therefore, the correct option is c.

A C-device should be used, which can maintain a nearly constant pressure throughout the infusion.

In both ab devices, the air pressure in the bottle will remain the same as the atmospheric pressure, but with the use of liquid, the liquid pressure will gradually decrease, so that the infusion speed will change.

The D device does not have an air inlet, and the pressure in the bottle will gradually decrease, and when it reaches a certain position, it will be balanced and the infusion cannot be continued.

The maximum current allowed by the whole galvanometer is certain, to change to a 3 voltmeter, it is necessary to make the total resistance after modification multiplied by the maximum current (IG) equal to 3 volts, so that the total resistance is 3V divided by 3mA is equal to 1000 ohms, so that the resistance value that needs to be connected in series is 1000-RG=900 ohms.

1. Since e=i(r+r), and e=blv, i=kt, kt(r+r)=blv. The solution gives v=kt(r+r) bl =at. where a=

k（r+r）/bl 。It can be seen that the metal rod ab moves with a uniform acceleration with an initial velocity of 0 in a uniform magnetic field.

2. Mgsin +F-pull-bil=ma is obtained by force analysis.

i.e. mgsin +(2)-bil=ma. i=e (r+r)=blv (r+r).

Mgsin +(2)-b(blv (r+r))l=mgsin + 2+(

Since the acceleration is a constant quantity and v is a variable, (must be 0

r= can be obtained by solution.

3. From mgsin + 2=ma, a=8 m s 2.

From s= , t= s is obtained.

Because f pull= , v=at=4 m s. Therefore, when the metal rod slides 1m, the average tensile force f=(f0+f4) 2=3(n).

From the transformation and conservation of energy we know fs+mgssin = , and we get q=fs+mgssin j.

The data is bothered to be checked again, okay? Yes or no, please refer to it!

I'll answer: b.

The resistance of the lamp that does not emit light is too small, and the series connection has the effect of voltage division, the voltage obtained by the large resistance is larger, and it can basically emit light, and the opposite resistance is smaller, and the voltage obtained is small, which is not enough to make it glow.

Hope it helps, thank you!

b in series, indicating good, and the current is the same. The resistance is small and the brightness is dark.

b p＝i*i*r

B does not emit light because the power is too small, it cannot be an open circuit, otherwise A does not emit light.

Because of the series connection, it is impossible for the filament to be broken, otherwise the circuit is broken, so A is wrong. Because of the series connection, the current is equal, so d is wrong. Because the voltage ratio is equal to the resistance ratio, and the current through the two bulbs is the same, so if the resistance is small, the voltage is small, the power is small, and it does not emit light, so C is wrong.

Therefore, choose B.

b Compared with the two lamps, the resistance of the bulb that does not emit light is too small.