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The tangent angle between the point P and the ground is 45 degrees, and the horizontal component of the velocity vp of the object at point P is equal to the vertical component.

VP Vertical Component = (2GR) = VP Horizontal Component = VD

vd==√（2gr =4 （m/s）

A, B exchange speed.

mgh=(1/2)m vd^2

h= (vd^2)/ 2g =r= (m)

3) The vertical height difference of the point m is r + rcos45° -r = rcos45°

1/2)m * vd^2 - mgrcos45°=m(gr - grcos45° )=mgr(1-cos45°) 0

So there is enough mechanical energy for the object to have the possibility to reach m.

If it can really reach point m, let the rate of arrival be vm, and the elastic force of the orbit to the object is n

1/2)m * vm^2 =mgr(1-cos45°)

mg +n =m vm^2 /r

n= 2mg(1-cos45°)-mg = mg(1-2cos45°)=mg(1-√2) <0

This means that the velocity is not enough to maintain the motion of the object along the orbit, so the object cannot reach the point m.

1) As a velocity vector triangle, it can be seen that the horizontal velocity of the block is equal to the vertical velocity at the P point.

vd = root number (2*g*r) = 4m s

2) The root and momentum theorem can calculate the velocity of block A when it leaves the chute: vama*va+mb*vb=mb*vd

2*va+0=2*4

va=4m/s

It is also followed by mgh = mv*v 2

There is h=v*v (2g)=4*4 20=

3) Unable to reach point m.

When the block is moving exactly to m, there is mg = mv * v rv = root number (g*r) = root number 8

Let the height of point d be 0, and the mechanical energy of the block will be mv*v 2+mgr=24>2*10*total mechanical energy).

The maximum tensile force is also the maximum centripetal force. So, f=

m×w×w×r=m

V VR (m is the ball mass, w is the angular velocity, v is the linear velocity, r is the radius). 5=

v×v/r25

w×w×rv×v/r

Look at the answer: a, one week per second, that is, w=1

Get r=25

m). b, v=1, then r=

m). That's right.

c、r=v=1。That's right.

d. 2 weeks per second, then w

2. Find r=25

4 = m). So choose B, C.

In this process, in the process of lifting three meters, the excavator overcomes the gravity of the soil to do work, w=fs=( and does not do work in the process of horizontal movement The number is calculated again to ensure accuracy.

The density of different substances is not the same, not the larger the mass, the greater the density, but also related to the volume; It is the ratio of mass to volume, and b and c are the same density of the same substance

I choose DA. For different substances, the greater the mass, the greater the density. The fault is that density is determined by the essential properties of matter. The greater the mass, the density can be smaller.

b.For the same substance, the density is inversely proportional to the volume. Wrong is the same substance, the density is the same, and the same is true for c.

All we can say is.

For the same substance, the mass is proportional to the volume, and the density is determined by the substance itself, and has nothing to do with the mass volume.

The first question is that the club does not exert force on the ball, and gravity provides centripetal force.

mg=mv2r, v=root gr

The second question is that in a uniform circular motion, the centripetal force of the ball must be the same, and the centripetal force of the ball at the lowest point is mg+

At the highest point f-mg= gives f=

, v=rg. under the root number

2.At the lowest point, the centripetal force f1=; Velocity of the ball: v1=(

Uniform linear motion.

Is it a linear motion with uniform variable speed?

When a train moves out of a railway station and moves in a straight line at a uniform speed.

Yes. Speed from 0

Begin. So let the acceleration a

The length of each carriage.

l then nl so n=25

Behind. Nine sections.

So let's do it first. First verse 16.

216lt'=16s

So. After filial piety is 9 knots.

20-16=4s

Do it with relative velocity.

in the second object.

When put down. v1=

g is taken because. is a multiple of 49.

Calculate it out. Probably better looking).

at any point in the future.

v2=gtso.

The relative velocity of the two objects remains the same.

Yes t = if.

g take 10m s2

So. t=still use relative velocity. ab

The acceleration of the is both.

Downward g so they.

between two objects.

The relative velocity remains the same.

So the time of the encounter t=

h v0 when they meet.

The amount of velocity change in b is. vgt

GH V0b meets the A-sphere during the ascent then δV V0 ie. gh/v0

v0 is the world and vo

The root number GH, B meet the A sphere during the fall then δv v0 i.e. GH V0V0 is VO

Root number ght = 120 (18-6) = 10s

Let's assume that the car has been.

Maintain acceleration to the right.

Then they met. Then there is.

18t''^2=6t'

t'=12s

That's the end of it.

That's wrong) v vapor = 18-2 * 12 = -6m s0m s but.

Cars are. Brake.

So when the velocity is reduced to 0 it is.

No further acceleration backwards.

Makes the speed change.

So. When they finally met.

The car is stationary.

The car brakes from the beginning to the standstill.

Distance lv 2 2a = 81m

t'=81/6=

Is it okay to do these processes?

The three balls are equal in volume, and now all three balls are floating or suspended in the water, and the gravitational force of the three balls is equivalent to their respective buoyant forces. Whereas, buoyancy f = density of liquid * g * volume of liquid discharged. The three spheres are equal in volume, only B is completely immersed in water, so it has the largest volume of water discharged, so B has the greatest mass.

As with the above equation, according to the buoyancy formula, it can be found that the buoyancy ratio of the two is 1:2, the gravitational ratio is also 1:2, so the mass ratio is 1:2, and the mass = density x volume, so the density ratio of the two is 1:4