Please help explain this physics problem, the answer is B

Because the volume is equal, so when he says density, he means that the mass of B is greater than that of A.

Pressure = weight divided by pressure area.

The pressure of the lower brick on the ground is equal to the sum of the two bricks A and B (which is not surprising), and the pressure of the upper brick on the lower brick is equal to his own weight.

Then it is only possible that the nail is on top (the nail is lighter. B is below.

Then in b, the contact area of B is large and the total pressure on the ground is large, and the contact surface of A above is small, and the pressure on B is small, according to the formula in the second row, it is "possible to be equal". As for the other why they are wrong, figure it out by yourself according to the formula, and only by thinking and thinking on your own can you improve! Come on!

The volume is the same, and the density is large B, then the mass of B is large, that is, the gravity of B is large. If the volume is the same, and the ratio of length, width and height is 4 to 2 to 1, then the shape of the two bricks should also be the same. a, the contact area between A and B is 2 times the width, and the contact area between B and the ground is the length multiplied by the width, then the latter is 2 times the former, if you want the same pressure, the gravity of A accounts for one-half of the total gravity of the two bricks, and it is known that the gravity of B is large, so A is impossible.

B, the contact area between A and B is wide times high, the contact interview between B and the ground is length times wide, the latter is four times that of the former, to be the same pressure, A's gravity accounts for one-fourth of the total gravity, B accounts for three-quarters, in line with the known gravity of B, so B is possible. C is the same as B, but B is on top, and B accounts for a quarter of the total gravity, which conflicts with B's gravity, so C is also impossible. d The two contact areas are the same, and the gravity is not the same, so it is also impossible.

Since i2 is an inductive current and i3 is a capacitive current, the two are opposite in phase, so the sum of the two is 8-5 = 3(a).

The resistive current at the i1 bit, and therefore the phase difference between the inductive and capacitive currents is 90°, thus: the total current i ((i1) +i2-i3).

5（a）

(1) The vibration equation is x=acos(t+).

So the amplitude a=

Angular frequency = 20Hz

Cycle = 2 20

The initial phase = 2) t = 2s is obtained by the formula.

Displacement x = velocity v = -a sin( t+ )2 sin( 4) = acceleration = f m =

There is no m value, and there is no indication of what type of simple harmonic vibration it is, so it is not calculated.

When the rheostat is connected in series with the lamp, and the slider moves to the A terminal, the total resistance decreases, the total voltage does not change, and the current in the circuit increases according to I=U R.

That is, the indication of the ammeter increases. The resistance of the lamp remains unchanged, according to the U lamp = IR lamp, it can be seen that the voltage at both ends of the lamp increases, and the voltage at both ends of the voltmeter measures the lamp, that is, the number of the voltmeter increases.

When P moves towards A, the rheostat resistance R becomes smaller and the series resistance becomes smaller according to i=e (R+R lamp+R).

The current becomes larger, and the degree of the voltmeter is U

According to e=u+ir+ir

Unchanged, i becomes larger and r becomes smaller.

Maybe the topic has a known condition to get it.

The answer should be C

In order to minimize the force, it should be maximized in the corresponding arm In this diagram, for the force acting at point C, the largest arm is oc. (Using the Pythagorean theorem, oc=10cm can be calculated).

f=g×oa/oc=30×10cm/10cm=30n<>

In order to balance the object, the resultant moment of the system to o should be 0, in order to minimize the force, the force arm should be the largest, and the maximum force arm at point c is co, so the answer is b, got it?

The work done by f w=fs is the same force, the same displacement, and the same work done. b correct.

The average power of f p=w t has a long motion time and a small average power when there is friction. C false.

The instantaneous power of f p=fvt is different at the end of the two cases, the instantaneous power is different, the end speedometer is large, the instantaneous power is large, and the terminal velocity is large and the instantaneous power is large in the case of frictionless. The instantaneous power in the latter case is high. D false.

You've made the right choice.

Please be prompt. If you have any questions, ask them in a timely manner. I'm here to help.

Analysis: AB option is to find work, constant force to do work is force f multiplied by the displacement that occurs in the direction of the force, w1=fs, w2=fs force and displacement formula is the same as the first time, so b is correct.

cd is the comparison power, c option is to compare the average power p average = fv average, because the force is the same, that is, the average velocity of the two comparisons, because the initial velocity is zero, but the same displacement occurs when the acceleration is different twice, and the final velocity of the acceleration on the smooth surface of the second time is large, and the average velocity is large, then p2 >is wrong.

Option d is to compare the instantaneous power, p instantaneous = fv instantaneous, f is the same to compare the instantaneous speed, the second terminal velocity is large, so the second instantaneous power is large, so p1'

The choice aw refers to work, which is defined as the magnitude of the force multiplied by the distance traveled in the direction of the force.

Then obviously, the magnitude of the force is the same, and the distance traveled is the same, then the definition of w1=f*s=w2 power is: the work done per unit of time. p=w t moves on a rough horizontal plane and is subjected to drag while on a smooth horizontal plane is not acted upon by drag.

t1>t2 then, p1t2 so p1=(w t1).

Solution: The work of the object on the rough surface: w1=1j Since the distance is the same and the force is the same, it can be obtained from w=fs, f is the same, s is the same, and the work of the object on the smooth surface: w2=1j i.e., w1=w2

The power of the object on the rough surface: p1=w1 t1 The power of the object on the smooth surface: p2=w2 t2 Since the force is the same, v1 v2, s is the same, so t1 t2 and because w1=w2, so p1 p2

Therefore, the answer is to choose b typing is very hard, hope.

w=fs, the constant force is the same, the distance is the same, so the work done is the same. Pick B

b. The correct work is only related to the force and the directional displacement product of the force, the rough effect is reduced, the resultant force is reduced, the acceleration is small, and the time taken to walk the same displacement is large, so the power is small

B in A is perpendicular to the surface of the paper, while V is on the plane of the paper, and the plane where V is located is perpendicular to B. In fact, as long as b, l, and v are at a 90° angle to each other, it is a three-dimensional scale calculation, so b is undoubtedly the correct answer.

You are wrong, the angle in A is the angle with the horizontal plane, not the angle with L or the magnetic field, it must be the angle with L or the direction of the magnetic field to multiply the angle.

In Figure A, the direction of the magnetic field is perpendicular to the inward surface of the paper, and the direction of the conductor cutting is parallel to the paper surface, so the conductor is perpendicular to the magnetic inductance line, of course, it is calculated with e=blv. The angle in the diagram is only the angle between the direction of movement of the conductor and the horizontal direction.