Sophomore electrical physics questions. 30 additional points in 1 hour.

1) R3 and C1 are connected in parallel at both ends of R2, which is equivalent to a voltmeter (because the capacitor is an open circuit), and the voltage at both ends of C1 is the voltage at both ends of R2 [U (R1+R2)]R2=4V, and the amount of electricity carried by Q1=C1U1= C

The voltage at both ends of C2 is the amount of electricity carried by U (C2 and R3 are connected in parallel at both ends of the power supply) Q2=C2U2=1*10 -5 C2) Treat R2, R3, and C1 as a circuit (in series) The power released by C1 is Q1 through R2

Think of C2, R3, R2, R1 as a circuit (series connection) where the amount of electricity released by C2 is Q2 through R2

So the amount of electricity passing through R2 is Q1+Q2= C

The voltage divider principle, the voltage at both ends of R2 is 4V. So the voltage across C1 is 4V. The voltage at both ends of the power supply is 10V. So the voltage at both ends of C2 is 10V. Calculate it yourself later.

After K is disconnected, the potential at the C1C22 terminal is exactly equal at the end. The amount of electricity that passes R2 is the sum of the charges of C1C2.

, U1 is the voltage at both ends of R2, U2 is the total voltage of R1, R2 in series, calculated, U1=4V, U2=10V, so, Q1=C1U1=4UF*4V, Q2=C2*U2=1UF*10V

After 2,K is disconnected, C1 and C2 are equivalent to two power supplies connected in parallel, forming two circuits: C1, R3, R2 and C2, R1, R2, R3, the power of each circuit has passed through R2, so the power that finally passes through R2 is Q1+Q2

Electrons do a flat-tossed motion in an electric field.

Vertical direction: t=lsin30° v0=5*10 (-9)s horizontal direction: lcos30°=1 2at 2

a= It's hard to type, do the math yourself.

a=qe/m

e = electrons are deflected to the right, so the direction of the electric field is in the -x direction.

(1) i=p u=(40+2*185) 220=fusing current i'=

2) The electrical energy consumed by the two motors per minute w=2p*t=44400j(3) The useful work done by the two motors per minute w'=w-i2rt=(4) The mechanical efficiency of this range hood = w' w=

1. The fusing current is the maximum current allowed to pass, and the current is the largest when it is just started, and the motor is treated as a pure resistance at this time.

The maximum current of each motor is i=u r220 90=, and the illumination current i=p u=40 220=, therefore, the fusing current i=(

2. The electric energy consumed by the motor per minute e=pt=185*3600j=666000j

The electric energy consumed by the two motors per minute w=2e=1332000j3, useful work = total work-thermal energy=w-q=w-i2rt*2i=p u=185 220

4. Efficiency is the useful work divided by the total work.

This question is to test the ability to comprehensively apply knowledge, when the maximum speed is reached, a=0, mg=bil, i=e r, e=blv, mg=....It can be solved v1, and this problem v1 is the solution r as a known number. The second idea is the same, but the relationship is different.

Finding q, since i is a variable, it is necessary to use integrals, which is beyond the ability of middle school students.

This kind of calculation question, who will answer it with a score of 0, really.

R1 is connected in parallel with R2, and the parallel resistance value is R(and)=(R1R2) (R1+R2)=and then connected in series with R3, and the total resistance is R(outside).

The output power P=IR(Outside)=12W, I=4, I=2AA, and the voltage between B is IR (and)=2

p = ui electromotive force u = p i = 15w 2a =

i r (inner) = 15w - 12w = 3w

r(inner)=3 4=

R1 is connected in parallel with R2, and its parallel resistance R'For: Ohm;

R3 is connected in parallel with R1 and R2;

The voltage between A and B is the voltage of R1 and R2;

r':r3=:1；

From the formula i 2r=p: i 2=p r=12 (,i=2;

AB inter-voltage u=;

Power electromotive force uelectric = p total i = 15 2 =;

Internal resistance r within = (p total - p out) i 2 = ohms.

Exactly the same metal balls are positively charged, after contact, divide the charge equally, first add the charge amount of A and B, and then divide Qi Rent 2, find that the current charge of each ball is 4*10 -9C, and then use Coulomb's law to calculate the shape of F2, so that you can know the relationship between the size or multiple of F1 and F2.

It's been said in great detail, please give it points.

Tips: 1. After two identical charged metal balls are in contact, the two small balls have the same charge, which is half of the original total amount (if it is a heterogeneous charge, it is good to neutralize the limb first and then divide the brigade type equally) 2.

Because of the fall, gravity does the positive work. According to the kinetic energy theorem, the magnitude of the velocity is unchanged, so the total work is zero, so the electric field force does negative work, so the electric potential energy increases. (The electric field force does positive work, and the electric potential energy increases; To do negative work, the potential energy decreases. Your teachers should have taught).

Let us set an equation and solve it relatively simply

The length of the carbon rod is x, and the corresponding gang is 1-x

Then find the temperature variable resistance of each of them around 0 degrees Celsius, and their sum is equal to 0