Help, urgently ask a physics question, help solve it, this physics problem, thank you

g=m*g=20*10=200n The gravitational force of the object is 200n and the drag force is: f1=g*

f-f1=100n-40n=60n

The acceleration of motion is the remaining force divided by the mass of the object a=60n 20kg=3n kgs=at》2 2=150m

w=fs=100n*150m=15000j, it's so hard

First of all, calculate the friction force f=40, so you can get 100-40=ma a=3 from Niu 2, and the speed is 30 after 10s, and you have walked 150 meters, so the work done by tensile force is kinetic energy plus friction force, which is 9000+6000=15000j

The 2nd floor is the right solution, the last step on the first floor, the 3rd floor, the two types are all wrong (I don't know if it's stupid or troublesome, it's really irresponsible, and it's wrong).

The acceleration of 3100N of horizontal tension minus 40N of friction divided by mass is 20 kg.

The work done by the pull f is 150

The horizontal tensile force of 100N is multiplied by the distance traveled in 10s.

Since i2 is an inductive current and i3 is a capacitive current, the two are opposite in phase, so the sum of the two is 8-5 = 3(a).

The resistive current at the i1 bit, and therefore the phase difference between the inductive and capacitive currents is 90°, thus: the total current i ((i1) +i2-i3).

5（a）

(1) The downward acceleration of the inclined plane is a=gsin -ugcos s=1 2at =1 2(gsin -ugcos) t (2) The horizontal acceleration of the block is a1=acos

Blocks are horizontally stressed ma1=macos =mcos (gsin -ugcos )

The force of the stake on the level of the inclined body is mcos (gsin -ugcos) (3)f (m+m)>=a1

f>=(m+m)(gsinθ-ugcosθ)cosθ

Answer: There are 3 collisions.

Solution] We don't care how the magnetic field makes m rest, in short, the whole process is: after the collision, m moves in the opposite direction, m deviates from the equilibrium position, and when m swings to the equilibrium position again, m is still in a resting state ,..And so on.

Let the velocity of m when it reaches the equilibrium position be v, and the collision occurs, let the velocity of m after the collision be v1 and the velocity of m v2.

Since it is an elastic collision, the momentum is conserved and the mechanical energy is conserved.

Column: Conservation of momentum: mv=mv1+mv2......Conservation of energy: mv 2 = mv1 2 + mv2 2......and is known to be m=19m......③

To solve the equation, substitute into , and then substitute into , approximate m, and solve v2=-9v1(It's a bit complicated, don't get it wrong) substitute the solution v2=

This equation is universal and applies to every collision. It shows that after each collision, the velocity of m is times that of before the collision.

At a 60° angle, when the angle of mgl(1-cos60°)=mv 245° is satisfied, the mgl(1-cos45°)=mv2 2 is divided by the two formulas, and v2 v =2-2=

So when m leaves the equilibrium position at a velocity that satisfies v2 v, the angle of deviation is less than 45°

Assuming that n collisions are required, then from the previous analysis, we know that v2 v= is substituted for "<".

The solution is n 3, that is, when the collision is 3 times, the angle of the swing will be less than 45°

Let the kinetic energy of the insulated ball reach its lowest point for the first time q

According to the principle of conservation of elastic collision energy, the velocity of the insulated ball after the first collision v1=((m-m) (m+m))v0=((m-19m) (m+19m))v0=

The kinetic energy after the first collision of the insulated ball is (, ie.

Each time the insulated ball is collided, the kinetic energy will be multiplied before the collision.

The potential energy at 60° is all converted into kinetic energy q, and when =45°, the potential energy is ((1-cos45°), approx.

is less than the third dimension, which means that the maximum angle at which the insulating ball deviates from the vertical direction after 3 collisions will be less than 45°.

Hello, that's the case, when the thin test tube can just "climb", the force of the thin test tube is balanced, and the water column above the test tube still has pressure on the small test tube inside, and the water pressure of the thin test tube is P0- 0GH0, because F0=G+F, so P0S=G+(P0- 0GH0)S

So h0 = 4g d2 0g

Hope it helps.

Only c. Analyse.

a。It can be calculated according to the formula of gravitation, which is obviously not true, and should be a formula related to the square of distance.

b。After the jet, the speed changes, the orbit also changes, and it is no longer a circular track, so it will not catch up.

c。Correct, according to the law of circular motion and the law of gravitation, the formula is easy to get.

d。The mass of the satellite is not known, and the mechanical energy cannot be determined.

A should be GR2R2

b acceleration changes orbit.

c should be: [(3+2k) r]* r 3 g)d The two satellites have the same mass, the same orbit, the same velocity field, and therefore the same mechanical energy.