Ask for a first year of high school physics, please ask me a year of high school physics .

This question asks for a rotational speed n

A is the first hole and B is the second hole.

While the bullet flies at point A turns to its current position.

The disc turns at an angle of 2k +2 3

k= 0 , 1 ,2 ,3...n minimum is 2 3 The flight time of the bullet is the same as the time of movement of the disc.

t = 2/400 = 1/200 s

Angular velocity: w = (2k +2 3) (1 200) w = 2 n

Rotational speed: n = w 2

2kπ +2π/3 /

200(k+1/3)

k= 0 , 1 ,2 ,3...n

According to v=s t, t=5*10 -3 can be obtained

And in time t, the angle at which the disk turns is k +2 3, k = 0, 1, 2 ......So we get w=k +2 , k = 0, 1, 2 ......According to t=2 w, t=3*10 -2 (k+2 3), k=0,1,2 ......

w=2*

But it has the potential to be more than one lap.

So he can spin 2 3 or 1+2 3 or 2+2 3 ...... per second

f=gmm r squared.

Coulomb force. The power of both Q and Q is E

Plausibly yes - to the power of 19.

Then you bring in the formula and you get the answer.

where g and k are constants. respectively. gxm

kgs2 k=

The Coulomb force is k*q1*q2 r r, k=9*10 9, and the gravitational force is g*m1*m2 r r.

2ax=v 2 so the magnitude of the acceleration a=1m s 2 net force f=ma=800000n

The magnitude of resistance f1=

So the magnitude of the braking force f2=f-f1=

The initial velocity is 20m s and the final velocity is 0 and the distance s = 200m the square of the initial velocity = -2as

A=Pro, just do the math.

The key to choosing a is the direction, e is the vector, what is the definition of the vector positive and negative on the coordinate axis? It is positive in the direction of the arrow, and negative in the opposite direction of the arrow.

The direction of e is the direction of the force on the positive charge in the electric field.

To the right of the charge on the right, the positive charge is stressed to the right, so it is positive;

Between the right charge and the origin, the positive charge is forced to the left, so it is negative;

And so on. In addition, the electric field at infinity is zero, and when the charge is close to the point, the electric field tends to infinity. So choose A.

aThe electric field strength is a vector quantity.

The closer to the field source, the denser the electric field lines, and the greater the value of e.

The electric field strength along the positive direction of x is positive, and vice versa.

b At the midpoint of the two charges, the electric field strength is very small, excluding d When the distance from the charge is 0, the electric field strength can be considered infinite. Therefore, choose B

ABCD, all wrong, but B is the closest, let the line of option B pass through point O.

1.The vector sum of the external forces of the system is zero, and the total momentum of the system remains the same, and the velocity of the sandbox is v1 at the moment when the bullet flies out of the sandbox

Conservation of momentum: 1 v1=

v1=4m/s

2.The mechanical energy of the box is conserved when it rises to the highest point: mgh 1 2 (mv 2) h=

3.It is known from the conservation of mechanical energy: the velocity of the box when it returns to its lowest point is 4m s.

f (centripetal force) t (tensile force of the rope) mg (gravity) t = mv 2 r + mg = 20n

You need to try to analyze the motion of the two objects yourself.

1 First of all, use momentum conservation to find the velocity of the sandbag, and subtract the final kinetic energy with the initial kinetic energy.

2. Conservation of mechanical energyWhen the kinetic energy is 0, the maximum height is reached.

3 At the lowest point, the pulling force minus the gravitational force is equal to the centripetal force.

That's the way it works.

First of all, according to the momentum theorem, we can find the velocity v (sandbox) = ( according to the energy theorem, and then find the energy loss w = (

So the loss is 1192j

The mechanical energy of the box is conserved when it rises to the highest point: mgh 1 2 (mv 2) h=

It is known from the conservation of mechanical energy: the velocity of the box when it returns to its lowest point is 4m s.

f (centripetal force) t (tensile force of the rope) mg (gravity) t = mv 2 r + mg = 20n

It is necessary to figure out what is the main test and what method is used to solve the problem.

The radius of the circumference r=lsin centripetal acceleration a= 2r= 2lsin

Linear velocity v= r= lsin

The radius of circular motion r=lsin

Centripetal a= r

Line v= r

For the sake of aspect, now on a planet, the gravitational attraction of the planet on the object is set to f, then f=g*mm r (r is the radius of the planet, m is the mass of the planet, and m is the mass of the object).

The object is on the equator of the planet, and it will also rotate with the planet, so it can be considered that the object is moving in a circle, and the radius of motion is r.

Since it is a circular motion, centripetal acceleration is required, and the direction of centripetal acceleration should be vertically downward, which is the same as the direction of gravity.

F-T1=MA=F-direction (T is the tension of the spring scale).

The actual reading of the spring scale is actually t1=f-ma.

But at the poles, you basically don't follow the rotation of the planets, and you can think that you are standing still, with t2=mg

According to the title t1=, i.e. f-ma =

Solution, a-direction =

The planet has a day and night of 6 hours, and we know that one rotation is one day and night, so we can find the angular velocity of the planet = 2 t

Again, we get a direction = r

Substitution in equal amounts. Now we require the average density, =m v=m (4 3 r)=3 (4 )*m r

The above formula can be simplified.

m/r³=10ω²/g=40π²/gt²

Based on the conclusion above, =3 (4 ).

40π²/gt²=30π²/gt²