High School Physics!! There are points plus !!

The ball originally had no kinetic energy. In the process of throwing the ball, the person has a strong effect on the ball and displaces the ball in the direction of f, that is, the person has done work on the ball. According to the kinetic energy theorem, the work done by a person on the ball when throwing the ball is equal to the kinetic energy gained by the ball.

Therefore w=ek=(1 2)mv. ^2

From the time when the shot is not made (the velocity is 0 at this time) to the highest point (the velocity is also 0 at this time) the kinetic energy theorem: w (man does work on the ball) -mgh-fh=0-0 to get w=mgh+fh (1).

From the time after the shot (speed is v0) to the highest point.

Kinetic energy theorem: mgh-fh=0-1 2 m(v0) 2 can be obtained mgh+fh=1 2 m(v0) 2 (2) from (1)(2) can be obtained: w=1 2 m(v0) 2

According to the kinetic energy theorem, the work done by the combined external force is equal to the change of kinetic energy, when the ball is thrown, because the force of the air on the ball can be canceled in all directions, only the person does work on the ball, and all of it is the kinetic energy of the ball from rest to v0.

The work done by the person on the ball is converted into the kinetic energy of the ball so it is mvo2 2,

All the work that man does on the ball is converted into kinetic energy, and all kinetic energy is converted into potential energy and heat, and there should be two answers.

1 2 mvo square and FH+mGH

What else is there and why ...

When A and B are thrown at the same height, because S A = 2S B, V A = 2V B. （h=1/2gt^2)。h is equal and time is equal.

s=vt., v A = 2v B. To s A = s B, t A = 1 2t B, h A = 1 4h B.

So the height of B is 4h

Twice as long.

Calculated according to the free fall formula.

Let t so when 2t is 2t, then multiply by 4=4h so it is 4h

I know, but there are no bounty points, too lazy to type.

This question is so easy! First, two people throw two balls from the same height h, and the horizontal distance of A is twice that of B, so the horizontal initial velocity of A is twice that of B (the same time in the vertical direction, 2h g is the same).

If you want the horizontal distance to be the same, you only need to move 2 times the original time, 2*4h g=2 2 to sum up 4h

Composed. Centripetal force m(w2)r=gravitational force gmm(r2)and. Period t=2丌 w

Get. r 3=gm(t 2) (4丌 2).

r1 r2)) 3=(t1 t2) 2=(1 8) 2=1 64 The ratio of the radius is.

r1:r2=1:4

v=wr=(2丌 t)r

The ratio of the speed is.

v1:v2=(r1/t1):(r2/t2)=(r1/r2)/t1/t2)=(1/4)/(1/8)=2:1

Since the ball is stationary at point a, it can be seen that the electric field force is horizontally to the right, negatively charged, and eq=mgtan53=4mg3

When there is a force of gravity equal at the pull force in C, the velocity at point C is 0

According to the kinetic energy theorem mgl-eql=-mvb 2 2, vb=(root number 6gl) 3 can be solved

Move to point D, the velocity of VD at point D is 53 degrees with the initial velocity, then the resultant force of the electric field force and gravity is constant force f = root number (mg) 2 + (eq) 2 = 5mg 3, f is at right angles to vd, then start from point d to do a flat throwing motion, x=2lcos53=vdt

2lsin53=ft^2/2m

The joint solution yields vd = (root number 3GL)2

You are the "Invisible Dream", I am the subject of the Invisible You! Can't help you......

Choose CD oblique split to move to the left, the heavy ball must swing to the left with the oblique split, and slide upwards on the inclined plane, the friction direction of the oblique split on the heavy ball is downward along the inclined surface, and there is a component to do positive work in the positive direction of the horizontal displacement of the heavy ball and negative work in the negative direction of vertical displacement, but in the end, the center of gravity of the heavy ball rises, and the gravitational potential energy increases, so the friction between mm does positive work on m, and ab is wrong.

Since the oblique split moves at a uniform speed under the action of F, the ground is smooth, and only the force of the heavy ball M on the oblique split is balanced with F, which is equal to the work done by F, and C is paired.

Because there is a relative sliding between the small ball and the oblique split, there is a pair of interacting elastic force and a pair of interacting friction force in the process respectively, because the displacement of the horizontal direction of the heavy ball (the heavy ball suspension line finally has a swing angle to the left) is less than the horizontal displacement of the oblique split, the vertical component of the heavy ball friction force does negative work, and there is no displacement in the vertical direction of the oblique split, and the vertical component of the two frictional force components is equal in size, so the friction force does different work on the heavy ball and the oblique split, and the sum of the work done by the support force and the frictional force is equal, so m, The absolute value of the work done by the elastic force between m and the work done on m is not equal, d pair.

It is the support force that does the work on m.

Friction does work on m and no work on m.

So A is right and B is wrong.

The elastic force between F and mm is unknown, and it is impossible to judge that the absolute value is not equal, and C is wrong.

d is the relationship between the action and reaction forces, and the absolute value is equal.

Option: a

I'll start with the first one.

l3=vt3+at3^2/2

l2=vt2+at2 2 2 l3 is the displacement of 3 seconds l2 is the displacement of 2 seconds v is the initial velocity t3 is 3 seconds t2 seconds.

l3-l2=

2*1+(9/2-4/2)a

a=1 second.

l/2=at^2/2=2a

So l=4a

l=at1^2/2

t=8 open squared=

pv=nrt

75+xx=Hum begging.

Hurry!! Ruined boy to make Hu.

The combined external force of gravity and the track on the train provides the centripetal force of the train turning, mv square r=mg tan@

tan@=h/r

When mv square r=mg tan@, the two rails do not put pressure on the side of the wheels along the turning radius.

MV square r is less than mg tan@ The outer ant slip grip rail has lateral pressure on the wheels along the direction of the turning semi-dull path.

When the MV square r is greater than mg tan@, the inner rail has lateral pressure on the wheel along the direction of the turning radius of the leakage.

1. When the particle does one-way linear motion, the distance and displacement are numerically equal, but the two concepts cannot be equated, the distance is a scalar quantity, and the displacement is a vector quantity.

2. Maximum displacement: 2r(m).

Maximum distance: 20 r(m) (approximately equal).

1.is wrong. When used as a straight shuttle motion, the displacement can be zero, but the distance is obviously not.

2.The maximum distance is, of course, 20*pai*r (10 circles in circumference).The maximum displacement is 2ris the diameter of the circle.