Elementary School Physics? Why the third question?

If the slider p is in the middle position, the length of the resistance line connected to the circuit by the sliding rheostat is half of the total length, so the resistance r is half of r; As can be seen from the figure, the voltmeter measures the voltage at both ends of R0, and when the slider p is at the position of the maximum resistance, the current in the circuit is.

U1 R0 When the slide p is in the middle position, the current in the circuit is .

U2 R0 sets the power supply voltage to U, then it is obtained from (1), and the power supply voltage U=U1+U1 R0

r is obtained by (2), and the supply voltage u=u2+

u2 r0 Synapse, r=

2(u2−u1)

2u1−u2

r0 If the resistance value of the sliding rheostat is too large, the voltage representation number will be too small, so a fixed value resistance should be added in series with the sliding rheostat, so the answer is: (2) length;

2(u2−u1)

2u1−u2

r0；too small; Fixed-value resistance

Looking at the landlord, I wish the landlord a happy new year!

The internal resistance of the ammeter is very small, and the internal resistance of the voltmeter is very small. The above connection method is equivalent to connecting the ammeter directly to the power supply in series (short circuit), so the current far exceeds the rated measurement range of the ammeter, and the ammeter burns out.

Question 2: Isn't the second null an ammeter--

The third case is to assume that the original amount of electricity is Q1 and Q2 respectively, and the distance between the charges is R1, and then put the second case Q1'=2q1，q2'=2q2，r'=r 2 to the formula of Coulomb's law, you can get the answer in the figure.

You didn't remember that formula yourself! The force of the charge in the vacuum, which is the product of two charges, divided by the square of their distance! The above two charges have become twice their original amount, so there are two bins multiplied.

That's four times as much. Plus the denominator where 1 2 squared. Isn't it just 1 4?

Turn it over again and multiply it. Isn't it 16 times?

This result is obtained by Coulomb's law, the original amount of charge is q1 and q2 respectively, when the distance is r, the force between them f=k times the q1q2 is more than the square of r, from the formula, the force between them is proportional to the product of the charge and inversely proportional to the square of the distance, after the charge is doubled, then the force becomes four, after the distance is halved, then r becomes the original 1 2, after the square is 1 4, then the four forty-six so it becomes the original 16 times.

The height of the remaining wooden blocks h=

The mass of the remaining wooden block m =

Kilogram. The buoyancy it experiences f = m g =

The volume of water it drains is.

v¹=f¹/pg=4/1000×10

The height of the upper surface above the water.

h2 = the volume of water drained from the wood block before a third is cut.

v=f/pg=

The total volume under the surface of the water in the container at this time.

v total = hs =

The volume of water in the container is.

v water = v total -v =

After cutting, the total volume under the surface of the water is.

v total * = v water + v =

The height of the water surface at this time.

h*=v*/s=

The upper surface is located from the bottom of the container.

h** = the original upper surface is high above the water surface.

h1 = the original upper surface from the bottom of the container.

h = the height of the upper surface descent h=

K discharges water, which leads to the decrease of the buoyancy received by B, which leads to the increase of the downward tension force of B to the right end of the bar, and the upward pull force of the left end of the bar to A, which in turn leads to the decrease of the pressure of A on the table. When it is reduced to zero, the tensile force of A is calculated by the force balance of A, the tensile force of B is calculated by the equal moment at both ends of the lever, the buoyancy is calculated by the balance of B, the drainage volume of B is calculated, and the drainage height is calculated.

The minimum indication of the ammeter = 6 50+

The maximum indication of the ammeter. Theoretically, when the slider slides to the far left, the ammeter will be short-circuited, the current will be quite large, and the ammeter will burn out. However, the rated current of the sliding rheostat is 2A, so when the sliding vane slides to a certain distance from the leftmost end, the current will reach 2A, and then slide to the left, the rheostat will heat up and even smoke, so the total current is 2+.

The maximum is, because the maximum passing current of the sliding rheostat is 2A, A1 that has, and I want to add it; The minimum current is the current when the resistance of the sliding rheostat is maximum, which is 6V 50, plus A1.

Because the motion state of the wire frame in the magnetic field is not known, so the energy conservation is directly used to solve the problem, from the beginning of the magnetic field into the wire frame to the line disturbance of the magnetic field: the gravitational potential energy is converted into the kinetic energy of the wire frame and the heat in the wire frame, so there is mgsin = (1 2) mv + q, and the speed of the magnetic field is asked.

Question 3 b Causes; The pressure of the liquid is the same in all directions, the density of the liquid is not the same, and the height of the two points is not the same.