I d like to ask you about physics, physics, physics

For teleportation, this is just a hypothesis at first. It doesn't exist in reality. Secondly, teleportation is not a jumping march as you say, but a time movement greater than the speed of light (to explain, the fastest speed now is the speed of light 3 * 10 8), he is not without a trajectory, but you can't see the trajectory of movement, so he still has a distance, which is the distance should be equal to the displacement, not less than the displacement.

In physics, is a numerical difference between the change of a physical quantity before and after the change of velocity from v1 = 1m s to v2 = 5m s, then the velocity change v=v2-v1=......4m/s

The time varies from 10s to 20s, t=10s

The temperature, from 30 to 50, increased by t=20

-Coefficient of friction.

m-Mass. g - acceleration due to gravity.

x- generally stands for displacement.

The maximum static friction f= mg, assuming that the sliding friction is equal to the maximum static friction, mgx can represent the work done by the frictional force when the displacement of the object is x.

Conservation of momentum calculates mass-to-velocity ratio, conservation of energy to find energy loss, and question 3 is to find the objectivity shift of the bullet (the frame of reference is now stationary to the ground.

The question will only be done if you do it yourself.

I won't write about conservation of anything.

mv=（m+m)v"

mvt=(m+m)v"t is mx=(m+m)x'

The ratio of x is 3:1 m:m+m=1:3 m:m=1:2, and the bullet velocity is one-third of the original.

1. The kinetic energy is one-ninth.

1. The total kinetic energy of the wooden block is one-third.

The reduction in kinetic energy is ninth.

8. The total energy is 450, and the current total kinetic energy is one-third, that is, 150, and the thermal energy is 450-150=300

The third question is to fix the wooden block, which is actually to find the opposite position of the bullet.

Bullet vs. position shift = Block vs. position shift + relative displacement = 3 + 1 = 4, you will feel better if you do it yourself.

Analysis: 1. Suppose the initial kinetic energy of the bullet is e0=1 2mv0 2, and the final kinetic energy of the bullet e1=1 2mv1 2, then.

e0-e1=1 2mv0 2-1 2mv1 2=400j Since the friction between the bullet and the wooden block belongs to the internal force, the action force is equal to the reaction force, so the horizontal net force is 0 and the momentum is conserved.

mv0=(m+m)v1

fl=1/2mv1^2

f(l+l)=1 2mv0 2-1 2mv1 2=400jSo, fl=100j

eu=400-100=300j

2mv0^2=400/

eu=1/2mv0^2-1/2(m+m)v1^2=300jm/m=2/1

3、fs=1/2mv0^2=450j

f(l+l)=f*4=400j, so, s=

(1) How much mechanical energy is converted into internal energy in this process?

Let the mass of the block m bullet m initial velocity v0 be v1 after the same friction force f1 2(mv0 2-mv1 2)=400=f*((m+m)v1=mv0

f* to solve the above equations.

1 2mv0 2-1 2(m+m)v1 2=300j(2)What is the ratio of the mass m of the block to the mass of the bullet m?

v0/v1=3/1

mv0=(m+m)v1

m∶m=2∶1

3) If the block is fixed, what is the depth at which the bullet penetrates the block?

h2 = centimeters.

Let the maximum velocity at the end of acceleration be v:

Acceleration time t1 = v a1 = v 2

Deceleration time t2 = v a2 = v 3

t1+t2=t=10

v/2+v/3=10

v=12m/s

Average velocity v'=v/2=6m/s

ab distance = v't=6*10=60m

Let the time from the lower end of the iron rod to the floor t, then the time from the upper end to the floor is t + according to h=1 2gt 2

1/2g(t+

2t+t=h1=1/2gt^2=1/2*10*

The height at which the iron rod falls from its lower end to that level.

It's all such a simple question. There is no technical content at all, a question in the first year of high school. I doubt you can get into your first year of high school, and you can't do simple math calculations. Really.

1, let the object acceleration time t1, deceleration time t2, then there is a1t1 = a2t2, t1 + t2 = 10, can get t1 = 6 seconds, t2 = 4 seconds, then the acceleration distance is s1 = 1 2a1t1t1, the deceleration distance is s2 = a1t1t2-1 2a2t2t2, a, b distance is 60 meters,

1.Or you can understand it this way: the relationship between the electric potential difference and the field strength is u=ed, if in the range of d, e is zero, then the electric potential difference on this road is u 0, that is, the electric potential is the same everywhere.

Of course, it can also be done from the perspective of the electric field force: w qu, of course, is also equal to fs, the field strength is zero, f 0, naturally can not do work, so w 0, q is generally not zero, so u 0, that is, the electric potential is the same everywhere).

2。The intensity of the electric field is generally expressed by the density of the electric field lines, and the direction of the electric field lines represents the change of electric potential, and the electric potential decreases the fastest along the direction of the electric field lines.

For example, an isolated negative charge has an electric field line diagram in which the field strength increases along the direction of the electric field line (along the electric field line to the charge edge, as can be seen from E kq r 2, r is strong at a small location).

I hope what I said helped you.

1. Because the electric potential decreases along the positive direction of the electric field line, the field strength is zero and there is no electric field line, so it does not decrease.

The electric potential is actually caused by the work done by the electric field force, and it would not change without the force.

2. False, because the relationship between the electric field line and the field strength is that the greater the density of the electric field line, the greater the field strength, so the electric field line density along the direction of the electric field line is not necessarily reduced, so the field strength may be larger and larger.

1.In a region with zero field strength, the electric potential is the same everywhere; According to u=e*d, e=0, u=0, so this statement is correct!

2.The electric field strength must be smaller and smaller along the direction of the electric field line, and the electric field generated by the negative charge, the electric field line points to the center and the denser the electric field line is towards the center, the greater the electric field strength, so this statement is wrong!

Set pilot mass m

The support force experienced by the pilot n=ma=8mg, upward.

The resultant force of the supporting force and the gravitational force is the centripetal force.

F-direction = 8 mg-mg = 7 mg = mv 2 r

v^2=70*800=56000

v = 40 times root 35