I am anxious to summarize the physics knowledge of the second year of high school! Senior 2 Physics

Choose C. You don't care about the nature of the circuit, just analyze it normally.

Did you learn about capacitors? It is just two parallel plates, and there is nothing in between, which is equivalent to an open circuit.

There is a capacitor on the left side of the voltmeter, open circuit, there is no voltage, and the number is 0

Although the voltmeter is often regarded as an open circuit, you must be clear about how it works. It's not that there is no current passing through the voltmeter, but its resistance is too large, and the current can be approximated as 0, but it is not actually 0

In other words, although the electrometer is connected to the voltmeter on the right, the right side is the path. The left and right are connected, which is equivalent to the power supply being directly connected to the electrometer, so the indication is 200, that is, the power supply voltage.

Answer: The capacitor is selected C, which can be understood as the two conductor plates are not in contact with each other, so the capacitor is open. When energized, one piece has a + on it and the other has a -, so to balance, the capacitor will be discharged. Similar to rechargeable batteries.

The capacitor acts as an open circuit, so the voltmeter becomes a wire and the electrometer measures the supply voltage. Therefore, the answer should be A

Use of Ohmic Meters:

1. Adjust the zero of the resistor first: short the two meters (direct contact), adjust R so that the pointer points to the scale of 0 ohm (far right).

At this point there is Ig E (R r).

2. After adjusting the zero point of the resistor, when measuring the resistance rx, the current is i e (r r rx).

rx has a one-to-one correspondence with i.

In this problem, while adjusting the zero point of the resistor, there is 300*10 (-6) (r r).

(r r) 5*10 3 euros 5,000 euros.

When measuring Rx, it is known by "pointer pointing at the dial positively", iIg2

So (r r rx) 2*(r r).

rx (r r) 5 thousand ohms (this resistance is also called the "median resistance").

If the electromotive force of the battery becomes smaller, the internal resistance becomes larger, but it can still be zeroed, indicating that Ig E1 (R R1) can still be satisfied, E1 is the reduced electromotive force, R1 is the internal resistance of the battery (increased), and R1 is the resistance of the strain rheostat.

Apparently there is (r1 r1) (r r) (because e1 e).

So at the same scale (with equal currents), the actual resistance value is smaller than the original resistance value, and the reading is still read according to the original scale indication value. That is, the result of this measurement will be (larger) compared to the result of the previous measurement.

i=e (rx+rg+r+r) ig=e (rg+r+r) pointer on the dial**.

i.e. i.e. i.e

i.e. Ig2=E (Rx+RG+R+R).

rx=rg+r+r=

No change

(1) Analysis of the force on the object: when the Lorentz force qvb is equal to the gravity mg of magnitude, that is, the support force is 0 at this time, and the object is not subjected to frictional force, the object moves in a straight line at a uniform speed, and the equation qvb=mg is listed, so as to find the velocity v=mg (qb), and then there is the definition of impulse i0=m*mg (qb).

2) i i0, the Lorentz force qvb is greater than the gravity mg, when the friction is subjected, the object decelerates, and the deceleration is reduced to the size of the first question, the object is not subject to friction, so finally the object does a uniform linear motion, and remains unchanged.

So by the kinetic energy theorem, only friction does the work in the whole process, so there is friction doing the work w e and finally -e starts.

According to the relationship between kinetic energy and momentum e=1 2mv 2=1 2(i 2) m

So w e final-e starts 1 2(i0 2) m-1 2(i 2) m (negative work).

Can you make your symbols clearer?

There are two situations: the car is overtaken in motion, and the car is caught up after it stops.

If case 1, then t 5, = > s = 4t = 7 + 10t - 2t 2, the solution is t = 7, which contradicts the hypothesis.

Case 2, s=7+10t - 2t 2,t=5=>s=32m,=>t=s 4=8 seconds, i.e. it takes 8 seconds to catch up with the car.

The bullet just passes through the plank", the key is the word "exactly", that is, the bullet has exhausted all the energy before it can just leave the plank, which is the energy exhausted, that is, the kinetic energy is 0, so the final velocity = 0, so the situation of "the plank is not fixed" can be ruled out. Because if the plank is not fixed, the plank has a certain velocity after the bullet impacts, and the bullet needs to have a certain velocity if it wants to leave the plank, which contradicts the bullet terminal velocity = 0.

Since the planks are all fixed, and the velocity of the bullet when it leaves the plank is 0, because the bullet moves evenly and decelerates, that is, a is constant, so the resistance of the plank to it is constant, and the kinetic energy of the bullet is all converted into resistance to do work.

Let the thickness of the board be d, the resistance to the bullet f, there is: fd=mv0 2 - mv1 2, fd=mv1 2 - 0, and the solution is: v1=( 2 2)v0.

Or according to the velocity-displacement formula, there are:

2a 2d=0-v0 ,2ad=v1 v0 , solution: v1=( 2 2)v0.

The process of deriving the velocity-displacement formula is given below

s=v0t+at²/2...a，v1=v0+at...b, by b=> t=(v1-v0) a....c, c is substituted for a to obtain: s=v0(v1-v0) a + a(v1-v0) (2a), simplified: 2as=v1 -v0.

The first type of question is a catch-up question, and there is an acceleration <0, which must be considered in the actual situation.

The second type of question can be read through the question, that is, a uniform deceleration linear motion, using the velocity-displacement formula. And here the mechanical energy is not conserved, but due to the conservation of energy, there is "all the kinetic energy of the bullet is converted into resistance to do work", which links the kinetic energy of the bullet with the work done by resistance, that is, the kinetic energy theorem is used.

If it is not the word "exactly", then it is necessary to consider the fixed plank, the plank is not fixed but adherent, and the plank is not fixed and not sticky, etc., the latter two cases need to use the momentum conservation formula, although the question is not enough, but these situations should be taken into account when receiving the problem. Have fun.

The first: let the time be t, then in the time to catch up with the car, the bicycle travels 4t meters, according to the distance formula, x=v0t+1 2at 2 Therefore, the car travels 10t-1 2·2t 2 4t=10t-1 2·2t 2 can solve t=6 seconds.

Question 2: According to the kinetic energy theorem, w=1 2mv 2-1 2mv0 2

Therefore, if the initial velocity is v0, then 1 2mv0 2=2·1 2mv 2 mv 2=1 2mv0 2 is solved to pass through the first plank, and the velocity v= 2 2·v0

c Capacitor For direct current, it is equivalent to an open circuit after the positive and negative poles are fully charged.

Our voltmeter is a big resistance in this circuit, relative to the capacitor, so the capacitor connected in parallel with the voltmeter can be directly removed, and the circuit properties remain unchanged.

The electrometer has similar properties to capacitance, and it is equivalent to an open circuit after being fully charged. Therefore, the voltage measured by the electrometer is actually the power supply.

This circuit has no current because it has a capacitor, and the capacitor is a straight through cross, so U'=200v

However, the electrometer connected to C cannot measure the voltage of the circuit, so u=0V

Answer: The velocity of 5 m s is composed of a horizontal velocity of 3 m s and a vertical velocity of 4 m s. By v 2-v0 2 = 2 g x 4 * 4 = 2 * 10x x =