Urgent.. Sophomore Physics. in strings, etc

Because of the smooth water platform.

So the friction between the plank and the water platform is negligible.

According to the law of conservation of momentum: m1v1 m2v2 m1v1 m2v2 is positive to the left, i.e. 1*200+1*0=-1*100+1*v2 can get v2 =

The speed of the block of copper when it moves to the rightmost end of the block is equal to the velocity of the block (in this case, the relative velocity of the block and the block is 0, so that it does not slide out of the right end).

f friction = m2g

Then there is the conservation of energy: m2v2 = (m2+m1)v total +f*s (frictional work) -1

In the column a displacement formula: v total -v2 =2as---2 acceleration a will calculate it yourself———3

The correlation to get the kinetic friction factor should be at least what it should be.

I haven't touched high school physics for a long time, and it may be a little forgotten and wrong, but you yourself are combining my thoughts.

Momentum is conserved. m(v1-v2)=m(v2-0)v2=

m(v2-v3)=m-plate(v3-0)--the end velocity of the plate and copper is the same.

2*v3=3m/s

V3=L=Scopper-S plate=(V2 2-V3 2) 2A-(V3-0) 2 2A=50cm

a=3^2-2*

f=ma=μmg

a/g=

The process is too complicated, so take your time. Answer:

Because"The negative charge of 2*10 -8C obtained when it hits the lower plate", you can go back from 2cm to 4cm from the lower plate. So eqd=mgh=2*10* so e=

So u=eq=10 9 2*10 -8c=20vc=q/u=

**Problematic message for me.

At the lowest point, it can be known from the knowledge of circular motion: Yuqing f-mg=mv r can be obtained from the conservation of energy: mgr=w+mv or chop 2 to get w 7mgr 8

Because 80% is absorbed, q=

Because: q=cmt

So t = q cm = degrees.

mg 2q, 30° electric field force = the horizontal component of the rope (i.e., mgtana) + the gravitational force between the two balls.

A is correct!

Analysis: Gravity is not taken into account. The intrashell field strength is zero.

Then the particle is not subject to external force in the process from O to A, and the kinetic energy remains unchanged. The cd option is incorrect. When leaving the charged body from point A, the electric field force in the early stage is larger, and the kinetic energy increases faster, that is, the slope of the graph line is larger than that in the later stage.

So a correct!

Solution: (1) The potential difference between two points of DA is ed=(mg q)l=mgl g.

2) Along the parallel to the inclined plane and perpendicular to the inclined plane, the force of the ball is orthogonally decomposed (figure omitted), according to the title, it can be obtained, the component of the resultant external force on the ball in the direction parallel to the inclined plane is mgsin53°-(mg q)qcos53°=,, then the component of the resultant acceleration of the ball in the direction parallel to the inclined plane is, let the velocity of the ball at point c be v, and 2(; In addition, according to the law of conservation of mechanical energy, 1 2mv = mg(2r). Combining the above two equations, we get: r=l 6.

3) The ball is subjected to the gravitational and electric field forces perpendicular to each other, and its acceleration in the horizontal direction is (mg q)q m=g, and its acceleration in the vertical direction is g. Then it moves to the left with a uniform acceleration linear motion with an initial velocity of 45° obliquely downward and zero initial velocity, and the combined acceleration a= 2g. Let the displacement from point d to point p be s, then solve the triangle (figure omitted), and we can get s 2 + (s 2) tan53° = 2r; In addition, according to the equation of motion, s=1 2at can be obtained, and the above three equations are combined to solve:

t = 2l (7g), i.e., t = [2l (7g)].