Two high school sophomore physics solutions. To the process. You can do just one question. Urgent. 5

A bullet with a mass of 10g is driven into the sandbox at a horizontal velocity of v0 50m s and then exits the sandbox at a velocity of v 10ms. (The bullet and the sand box have a very short working time) g takes 10m s2. Seeking:

1) How fast does the sandbox get when the bullet flies out? (2) What is the maximum height of the sand box after gaining speed? (3) When the sand box returns from the highest point to the lowest point, how much is the pull force of the rope on the sand box?

Solution (1) When the bullet flies out, the sandbox obtains velocity v The mass of the bullet is m=10g, and the mass of the sandbox is m=1kg, according to the law of conservation of momentum mv0=mv+mv so v=(mv0-mv) m= The velocity of the sandbox when the bullet flies out is.

2) After the sand box obtains the velocity, the maximum height of the ascent is h, according to the law of conservation of mechanical energy mv2 2=mgh h=

3) When the sand box returns from the highest point to the lowest point, the tensile force of the rope on the sand box is f, and it can be seen from (2) that the sand box returns from the highest point to the lowest point is still v= This can be obtained from Newton's second law f-mg=mv2 l rope length l=so f=

Your order of magnitude is gone, so you can only be given the formula and the approximate process, and when calculating, you have to correct the data.

1) The total resistance of the transmission line is.

r= l s=5 10 ·m*2*400km 8cm =2) The power lost by the line.

p-line = i*i*r=

So the current. i= (p-line r).

The total power is W

So the delivery voltage is u = p total i

So. The conveyor ratio is 2000V:U=

3) The voltage drop on the line is:

U line = i*r

So the voltage at both ends of the original coil is u=u-u-u.

2.(1) According to the law of conservation of momentum.

m sand v0 = m sand v1 + v

2) According to the law of conservation of energy.

mv*v/2=mgh

3) When the sand box returns from the highest point to the lowest point, the tension of the rope on the sand box.

f=g+mv*v/r

r is the length of the rope.

Since this coil is rectangular, the area of the coil subjected to the magnetic field lines is changing as the coil rotates, so the magnetic flux is changing, not not not changing (i.e., the change is zero). Assuming that this coil is circular, then no matter how the coil rotates, the area of the coil bearing the magnetic field lines is constant, and then the magnetic flux is also constant.

Key points: 1. The coil is rectangular; 2. Zero change means that there is no change at all, not a cycle from large to small and then to large.

For example, consider the change in magnetic flux that passes 90° from the initial position.

First of all, we must understand the definition of magnetic flux: magnetic flux is the number of magnetic inductance lines passing through a certain surface, and in a uniform magnetic field, the relationship between magnetic flux and magnetic induction intensity is: =bssin ( is the angle between the magnetic field b and the coil s).

1. Magnetic flux at the initial position: 1=bssin0°=0.

2. The magnetic flux after 90° rotation is: 2=BSSIN90°=BSs.

3. The change of magnetic flux in this process is: 2- 1=bs-0=bs

The reality is that the magnetic flux passing through the coil plane changes the most for each half turn of the coil plane. So the choice d is wrong.

The correct option is A. Because the coil plane rotates at a uniform speed according to the OO axis, and the coil plane is perpendicular to the magnetic field lines in the initial state, the current intensity changes with time and presents a cosine change state.

Half a turn is 180°, and the magnetic flux passing through the coil plane is the same but opposite in the direction compared with the two positions at the beginning and the end. Therefore, the amount of change in magnetic flux is twice that of the initial position.

Each half revolution of the coil plane is a process in which the angle between the coil plane and the magnetic field is from 0° to 90° to 180°. Except for 0° and 180°, the magnetic flux at all angles crosses the coil plane, and gradually increases, reaching a maximum (at 90°) and then decreasing.

This process is contrary to the conclusion given by option d.

I don't think D is right or C, can you tell me why C is right, when the coil plane is parallel to the magnetic field, isn't it time for the current in the coil to change direction, and when the current is zero, the induced electromotive force should also be zero. c. If you change it to vertical, it would be right.

For the sake of convenience, let the angle between the coil plane and the magnetic field be . Then, =t the magnetic flux through the coil at this time:

nbs * sin = nbs * sin( t) is the induced electromotive force at this time by definition.

e = dφ/dt = nbs * d[sin(ω t)]/dtωnbs * cos(ω t)

As can be seen from the above equation, when t = k * k z, cos = 1, the induced electromotive force in the coil has a maximum value e = nbs.

So, only answer c is correct.

Note: The key point is that the induced electromotive force is the rate of change of the magnetic flux, not the actual value of the magnetic flux.

The correct answer is D, the coil rotates half a turn is 180 degrees, at this time the positions of AB and CD are exactly interchangeable, no matter what position the coil plane is in, the magnetic flux will not change, the original amount, and now how much.

If the coil plane is rotated once, the change is equal to zero, and now the title says that it is half a turn, which should be equal to NBS.

Every time the coil turns half a turn, the magnetic field lines passing through the coil change from the maximum to the smallest (or the opposite), and the rate of change of the magnetic flux is not zero.

The first horizontal direction is to the left, (2)v=2m s

3) The result is:

Well, this is our monthly exam question.

Before I do it, let's do a little analysis:

1.The object is subjected to gravity and electric field forces, and the acceleration is uncertain due to the uncertainty of the magnitude of the electric field force. Since it is released from a standstill at point A to a stationary point to a stationary point B, the object accelerates first and then decelerates, that is to say, the acceleration at point A is vertically downward, and the acceleration at point B is vertically upward.

Solved with Newton's second law f=ma).

2.Only the work done by the electric field force from a to b is required, and then it can be solved by u=w q, i.e., w=uq.

The answer upstairs is not known in M.

1) When the point charge mass is m and the point charge is from a, the force analysis: by gravity and Coulomb force, the resultant force is.

f = mg-f coulomb force a = ma = 3 mg 4, f coulomb force a = mg 4 let the distance between the two charges be r, f coulomb force 1 r

f Coulomb force a f Coulomb force b = (

F Coulomb force b = 16 F Coulomb force A = 4 mg

So the acceleration of the point charge at b is (4mg-mg) m=3g, and the direction is vertically upward.

2) At point A, UAQ = Mg4 UA=Mg 4q at point B, UBQ = 4 mg UB = 4 mg Q

The potential difference between the two points UAB = UB-UA = 15 mg 4q

First of all, the first space, your R1 and R2 in parallel after the resistance value is very small, approximately, so that the ammeter will be directly connected to the battery in series, although this problem will not burn out the ammeter, but in electricity can not see the ammeter and the battery in series, so your method is not advisable.

The second void sees**, you are wrong in the first void, and the second void is definitely wrong.

The second question, well, I am opportunistic, I have done this problem, I have done it first, I have skipped this option, and when I do the third empty, when I calculate the voltage of the battery, the internal resistance is 50 ohms, and then I will do the first short, assuming that the resistance of the rheostat is now 0 ohms.

Because the maximum current is 50mA, the resistance is, minus the internal resistance is 140, so close to option B, you can also look at the electric power, which is almost the option B.

Title.

1. The answer given in the book is correct. Upstairs said that "the internal resistance of the battery can only be ignored to have a solution", which is not right.

Set: the electromotive force is E, the internal resistance of the ammeter is RG, and the internal resistance of the battery is R3

For the first time, connect R1, ammeter, battery, and electric button into a series circuit, close the electric button, and measure the current i1;

e/i1=r1+rg+r3 --1

The second time, connect R1, R2, ammeter, battery, and electric bonds into a series circuit, close the electric bonds, and measure the current i2;

e/i2=r1+r2+rg+r3 --2

At the time of the test, the electromotive force of the battery does not change, i1, i2 are the measured values, r1, r2 are known quantities, and the 1,2 formula can be obtained by synthesis.

The electromotive force of the battery e=i1i2r2 (i1-i2).

The second question, the practice on the first floor is very good. For such a question, skip this option first, calculate the rest, and then think about it clearly.

Analysis: (1) Simplify the circuit to know that 3 and 6 resistors are connected in parallel with the resistors in series, 5 resistors are connected in parallel with their equivalent resistors, and the ammeter measures the sum of the current of 5 resistors and the current of 6 resistors. The equivalent resistance of 3 and 6 resistors in parallel is 2.

So, 5 resistors are connected in parallel with 20 (2 +8 +10). 6. The voltage at both ends of the resistor is 18 20 2V. So the current representation should be 18 5A+

2。The answer is right. The power supply voltage can be set to U, with UR2 (R1+R2)=60V, U=3R1; The key is the following analysis:

Since R3 is connected in parallel at both ends of R2, the power consumption of the three resistors is equal, so R2=R3, the current through R1 is twice that of R2, there is a power formula P=i R to get R2=4R1, and substituting the data to solve the equation system can get U=75V, R1=25, R2=R3=100.

My mailbox.

The low wind and cold groaning are correct.

I would like to add to the second question of the first question.

When connected to the voltmeter, the circuit is: 10 and 8 in series for 18, 5 and 6 in series for 11, 18 and 11 in parallel, and then in series with 3.

You first calculate the current A through 3 and the current B through 6 respectively, and the voltmeter shows 3 A+6 B

The topic is not complete!

u = 1:5

u = 2000v

2. Is it 220 ohms?

Total current i = u r = 2000 220 = power p dissipated on the resistor'= i*i*r = 18218w = Quite inefficient.

n = (p - p')/p =