When it comes to junior high school physics questions, the master comes, and junior high school phys

b, the principle of leverage, m A g * L A = m B g * L B; Because on the whole, the center of gravity of the rod must be closer to the first end, that is, the fulcrum (the place where the rope is hanged) is closer to the first end, so there is a l armor< L B, you can get M A > M B.

m A l A = m B l B.

l A M A M B .

Moment balance is junior high school knowledge.

l1g1=l2g2

L1 is shorter than L2, so G1 is larger than G2.

L1 is the distance from the center of gravity of section A to the rope, and L2 is the distance from the center of gravity of section B to the rope.

G1 is the gravitational force of segment A and g2 is the gravitational force of segment B.

With the principle of leverage.

According to the power * power arm = resistance * resistance arm.

It can be thought of as gravity concentrated at both ends, with a small mass in the middle (the ideal state in secondary school physics is considered 0), and the distance from the first end to the rope (i.e., the fulcrum) is less than the distance from the end to the rope.

After truncation, the mass of the nail end is still greater than that of the left end.

It can be considered in this way:

From the knowledge of moment balance, it is known that in order to maintain balance, the force on both sides of the stick must be equal, you can set.

If the mass of segment A is m, the mass of segment B is m, the moment of segment A is l, and the moment of segment B is l, then there is:

m*l = m*l, because l l, so m m

So choose B

Let the mass of section A be m, and the mass of section B be m, which is balanced by the moment: m*l=m*l, because l l, so m m b pair.

It is solved by power = force x speed: traction force f = 4800kw 200km h = 4800x3600 200 n = 86400n and driving at a constant speed then the force is balanced in the horizontal direction, and the traction force does work w = 86400x100km = 86400x100x1000 = j

The drag force is equal to the traction force f = 86400N

Because the car is moving at a constant speed, the combined external force is zero, that is, the traction force is equal to the friction resistance p = f * v, and the traction force f = 4800kw 200km h = 86400 N work done = 86400 * 100 * 1000 = 86400000 joule resistance = 86400 Nm.

p=4800kw

v=200km/h

s=100km

The work done is: w=pt=ps v=4800*10 3*(100 200)*3600=

Resistance received: f=f=p v=4800*10 3 (200

Because the train moves in a straight line at a uniform speed, the traction force and resistance are a pair of balanced forces (the drag force is equal to the traction force).

That is: f=f and because p=w t=fs t=fv, there is:

200km/h=200/

f=p/v=4800000w/

So there is resistance: f=f=86400n

Work done: w=fs=86400n*100000m=8640000000j

The gravity of this ruler is 10N, which is a lever problem, O is the fulcrum, and the gravity of the ruler can be set to g, and the ruler length is L, then there is: 2 3g times 1 3L = 1 3g times 1 6L + 5N times 1 3L, so g = 10N

f1l1=f2l2）

Set the length of the ruler l, the weight m new, m * 1 6l = 5 * 1 3l, m = 10 nN.

Those two questions are correct.

Let the gravity of the ruler be g and the length be l, and the center of gravity of each part of the ruler divided into two parts by the edge of the table is at the midpoint of each part, so there is:

2g 3)*(l 3) = (g 3) (l 6) + 5n * (l 3) solution gets: g = 10n

Note: If there are more than one force acting on the lever, the formula for the equilibrium condition of the lever is:

f1l1+f2l2+..fnln=f1l1+f2l2+..fnln

Also, the direction and application point of the force in the diagram you made are correct.

10n We have the original question ...... in our workbookIt's hard to fart.

**The law of combination is to trade-off the other.

If the aperture is 10 to 5, it is 1 stop larger (the larger the aperture, the smaller the value, and the larger the aperture, the larger the amount of light enters), the corresponding shutter speed should be smaller, and the shutter speed should be 125 from 15 backwards, and the 1 stop is 125 (the further back the shutter value, the faster the speed, the smaller the amount of light entering).

The shutter speed should be 1 125s, i.e. the ** combination is f 125

10 is the diameter of 1 10, and 15 is 1 15

Multiply the square of 1 10 by 1 15 and divide by the square of 1 5 to get 1 60

The shutter is 60

Solution: (1) Suppose that the volume of the extracted unit brine is v cubic meters.

Then the extraction brine mass m(ys) = brine v kg

We think of the extracted brine as adding salt to v cubic meters of water (if the salt is completely dissolved in water, the change in water volume is negligible).

v cubic meter water mass m (s) = water v kg

So the mass of salt m(y) = m(ys)-m(s) = (brine - water) v kg The mass of salt in brine per cubic meter t = (brine - water) v v=) = brine - water.

It can be said that salt is useless.

Unless you want to take into account the volume of salt dissolved in water.

then the system of equations is solved.

Salt v salt + water v water = brine v

v salt + v water = v

Because we assume that v is known to be a system of binary linear equations, the two sides of the equation are multiplied by water and subtracted.

The solution yields v salt = ( saline - water) v ( salt - water) so t = v salt * salt v = salt ( salt - water) v ( salt - water) 2) p brine = (p water + t) (1 + t p salt) 1000ml = 10 -3m 3

From the above formula: solution t=

It shouldn't be the third year of junior high school, and I haven't studied these questions.