Physics problems that you can t do, ask for help, physics problems that you can t do

The speed of the car is 10m s, and after 1s, the speed is 6m s, so the average speed during this time is 8m s, so the distance in this second is 8m

After passing by, the car does not move, and the last car does not move. The average speed of this is 5m s, so the distance of this is.

If you have any questions, please ask. Hope.

x=vt+

v=10m/s,a=﹣4m/s²,t=1s.then x=8m, and the car needs to slow down to 0m s, then the braking distance is.

First convert to 10m s, and then.

1s speed is reduced to 0 and stops, so.

s=vot-1 2at 2, set of formulas, s=8m has stopped, you can't use the above formula, use v 2=2as, get s= happy learning

Solution: v0 = 36 km h = 10 m s

x1 = v0t1 + 1 2 at1 = 10 1 + (-4) 1 2 = 8mv stop = v0 + at stop = 10-4 t = 0 t stop = and t2 t stop.

, so x2=(vstop -v0) 2 (-4)=

1. 1 The buoyancy of an object in water is equal to the force of expelling water, and the buoyancy in salt water is equal to its own gravity (levitation).

2 The volume of the object is equal to the volume of drained water, cubic meters.

3. When the object is suspended, the density of the salt water is equal to the density of the object = the mass of the object Volume = cubic meters = 1100kg m3

2. The pressure of 1 small truck on the ground p=n s=g s = 5800*g 2 The size of the resistance of the small truck is equal to the traction force of 4000N (constant speed).

Solution: Let the masses of an iron ball and an aluminum ball be a and b. respectively

From the inscription, 3a = 6b

So m:n=2:1

According to m=pv, v

Iron. =a/p,v

Aluminium. =b p, which brings the density of iron and aluminum into this formula. V iron: V aluminum.

It's very simple.

mgh=cmδt

h=cδt/g

Physics problems generally need to find a connection from the known, and they can be solved quickly.

No picture, according to your description, it should be A on B and B on the ground.

A and B do not move, and the friction should be static friction.

1.The frictional force of the ground facing B, taking A and B as a whole, and combining F1 and F2, the friction force is opposite to the same magnitude of the resultant force. The meaning of the title f1 and f2 is the same, the resultant force should be 7 New, the direction is the same as f1, the friction force should be 7 new, and the direction is opposite to f1.

The friction force against A and the tensile force on A are a pair of equilibrium forces, so the magnitude of this friction force is the same as that of A, and the direction is opposite.

There is no limb diagram, if f1 pulls a, f2 pulls b and goes in opposite directions.

1.The friction force is 7N, and the direction of the return is opposite to that of f2.

Contrary to f1.

foay=foa*sin60 degrees 150n*foby=fob*sin30 degrees 100n*According to the principle of equilibrium, the tensile force of the rope OC on the object should be balanced with the resultant force of the two components of foay and foby.

g=foc`=foay+ foby=

That is, the weight of the suspended weight must not be exceeded.

Hope it helps.

Suppose oa exactly breaks fob=foa times tan30* fob=50 with 3 < 100n, i.e. the rope ob will not break.

The weight g=100 with 3 < 200n rope OC will not break.

Suppose the OB rope happens to break, FOA=100%Tan30*=100 with 3>150N, i.e. the rope OA breaks.

Suppose the weight of the object is 200N, at which point the rope oc happens to break, FOB = 200N times sin30 = 100N

FOA = 200N times CO30* = 100 with 3 > 150N i.e. rope OA breakage.

In summary, the weight of the object should not exceed 100 and 3n

Let the weight of the object be m, then the force borne by OC is mg, the force borne by OB is 1 2mg, and the force borne by OA is three mg of the root number of one-half, and with the increase of m, it is easy to obtain the maximum tensile force of OA first, therefore: the root number of two-part of the root number is three mg = 150N, and it is calculated that M is equal to 10 * three kg of root number.