Physics experts can help!! Thank you very much

Hello, first of all, the velocity v1 is decomposed in the direction of 60 degrees into the velocity v0=2 root number 3 perpendicular to the river bank and the velocity v3=2 parallel to the river bank, plus the river velocity v2=3, so the horizontal velocity is v=2+3=5, and v and v0 are synthesized again to get v'= root number 37The shortest distance is along the perpendicular to the river bank, and its vertical partial velocity is the hydrostatic velocity of the ship 4 m sTime 25 seconds.

The shortest distance to cross the river is the distance to the speed of more than 25 seconds: 2 times the root number 5 times 25 seconds, which is 50 times the root number 5

If the velocity of the water is 5m s and the velocity is root number 41, the distance is 25 times that of root number 41

You're a freshman in high school, right? Only then did I learn the problem of crossing the river by boat, and the progress was a bit slow. The first question, force analysis, the resultant force can be vertical.

The second question is that the time is the shortest, so that the bow of the boat is perpendicular to the river bank, so that the direction of the river flow is perpendicular to the direction of the boat's movement, then the flowing water does not do work, no matter how much water flows, the time is equal to the time required to cross the river vertically when there is no water flow. It turns out that the current does not help to cross the river in any way, no matter where the bow of the boat is oblique, orthogonal decomposition shows that the component perpendicular to the bank of the river is reduced, so the river can only be crossed vertically. Got it?

Kiss???

1) The bow of the ship is at an angle of 30 degrees to the direction of the perpendicular bank (opposite to the direction of the current), which can be obtained by the parallelogram rule.

2) The shortest driving time of the vertical river bank is t=20s, x=3*20=60m, and 60m downstream of the riverbank

There is one less condition for the question: the electron enters a uniform magnetic field perpendicularly.

Solution: First find the radius of the electron trajectory, which is obtained by q*v0*b m*v0 2 r and the radius is r m*v0 (qb).

1) From the knowledge of triangles, (op 2) r sin is obtained, and the length is op 2r*sin 2*m*v0 *sin (qb).

2) Since the direction of the magnetic field is not given, there are two possible time outcomes.

Probably 1, the trajectory from O to P is less than half a circumference, because the direction of the electron velocity changes 2, so.

The time obtained from the period t 2 m (qb) is t (2 2 )*t 2 *m (qb) (here the unit is radian).

Probably 2, the trajectory from o to p is greater than half a circle, then the direction of velocity changes (2 2), so the time sought is.

t ＝[2π－2θ）/（2π）]t＝2（π－m /(qb)

1) The average velocity of the board acceleration phase v'=v/2

The displacement of the plate s=v't=vt/2

The displacement of the blocks'=vt

Relative displacement l 2 = s'-s=vt/2

So the displacement of the plate s=l 2

According to the kinetic energy theorem: (1 2) mv 2 = coefficient of friction * mgs so coefficient of friction = mv 2 mgl

2) The first three lines of the first question still apply.

When the velocity of the plate to the rightmost end is exactly v, the coefficient of friction (2) is the smallest.

Relative displacement l=s'-s=vt/2

So the displacement of the plate s=l

According to the kinetic energy theorem: (1 2) MV 2 = coefficient of friction * mgl - coefficient of friction (2) * (m + m) gl

Friction coefficient (2) min=v 2 2gl

When the two coefficients of friction are equal, the coefficient of friction between the ground and the plate is the largest.

coefficient of friction (2) max=v 2 gl

3) The displacement of the block s2 = l + l = 2l (the formula 3, 4, 5 of the first question is obtained) t2 = s2 v = 2l v

Tensile force f=i t2=mv t2=mv 2 2lw=f*s2=mv 2

END It might be easier to use blocks as a frame of reference, so you can try it.

1. Because the rope pulls the block at a constant speed, the desktop is smooth, and the block can only reach the middle of the plank, which means that after the block reaches the middle of the plank, the two are relatively stationary and move together at a constant speed of V. From the beginning of the movement to the time the block reaches the middle of the template, the block walks 2 more than the board, there is:

vt-l/2=

t=v/aa=μmg/m

Solution: =mv lmg

The displacement of the plate is s=l 2

2. When the block can just reach the right end of the board, the friction factor between the plate and the desktop is the smallest, and the plate is subjected to the friction of the block to the right and the ground to the left

vt'-l='t'²

t'=v/a'

a’=（mg-μ'm+m)g) m ( already obtained in the first question) gives '=mv 2l(m+m)g

So ' MV 2L(M+M)G

3. In this process, the relative motion friction between the block and the plate is done, q1=mv;

The work done by the frictional force of the relative motion of the plate and the ground is, q2=mv ;

Conservation by energy, there are:

W-Q1-Q2 = E Matter + E Plate.

Therefore w=(mv +5mv)2

Simple force analysis problem.

Ideas. In the question, the oblique body is subjected to the friction of the ground.

Because the bevel body is stationary.

Balanced forces. We can start with a force analysis of the inclined body.

Force on the inclined body:

1.Gravity supports the force (vertical).

The pressure on it (in the direction of the vertical bevel).

3.The friction of the ground (horizontal).

Hypogonides are subjected to only these three forces.

Because the friction force of the ground on it is horizontal.

So we only need to analyze the force in the horizontal direction.

Break down the pressure of a on it.

It is decomposed into vertical direction and water converse direction.

then the horizontal component is balanced by the frictional force on the ground.

The sizes are equal and the direction is reversed.

Seems too verbose.

(1) The pressure ratio of d is 7:6, then it is reduced by 1 7, and the upward pulling force is: f2=g*1 7=1395n*1 7=195n

f1l1=f2l2 f1==f2l2/f1=195n*4/3=260n

g + g wheel = 260n g = 260n - 60n * 2 = 140n (2 pulleys).

W has W total = GH ( G + G round) H = G (G + G round) = 140N (140N + 60N) = 70%.

2）p=g/s s=g/p p1=(g-f)/s=g/s-f/s=p-f/s

f=(p-p1)s=(p-p1)(g/p)=(

p=w t=fs t=fv=200n* (rope is 3 sections).

It looks complicated, but it's actually relatively simple, and you have to read it carefully to understand it.

From the pressure formula, it is known that for every meter that the water rises, the pressure on the pipe is.

Increase p = gh = 1000 * 10 * 1 = 10000 Pa.

Liquid pressure is related to the density and depth of the liquid, and is not related to other factors such as pipe diameter. According to the liquid pressure formula p= gh, the pressure on the pipe increases for every meter of water elevation p= gh=1000kg m 3 10n kg 1m=10000pa

<> perigee gravitational force is perpendicular to the direction of velocity equal to centripetal force, calculate the ratio of centripetal force to mass and instant centripetal acceleration.

2 Calculate the volume divided by mass to get the density.

Find f [Solution] When b is on a, b accelerates the motion, the acceleration a1=ug=1, the initial velocity of the deceleration at the end of acceleration is vt set the acceleration process, a relative to the desktop displacement sa1, in the acceleration process, the displacement sa2, then.

Analysis: (1) The heat released can be calculated by the formula of the combustion value (2) The power p===fv, and the resistance of the car can be found by applying the deformation formula of p=fv

3) Find the energy released by the gasoline engine to consume gasoline, find the generated electrical energy by w=uit, and then find the efficiency of the gasoline engine

Answer: Solution: (1) The heat released by the complete combustion of gasoline q discharge = mq gasoline = 2;

2) v = 72km h = 20m s, p = = = fv, the traction force of the car f = = = 600n, the car moves at a constant speed, the traction force and the resistance are balanced, and the resistance is obtained by the balance condition: drag f