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Momentum is conserved first, because the orbit is in a smooth plane: m is the mass of the orbit, and v1 is the final velocity of the small slide at rest relative to the orbit (for ground motion).

mv0=(m+m)v1, and then form a system of equations 1 2mv0*v0=8 (initial kinetic energy 8 joules).

v1 = 1 meter per second, v0 = 4 meters per second.

That is, the work done by the frictional force in the relative motion between two objects is equal to the reduction of the mechanical energy of the system. So wf=ek2-ek1

wf=(1/2)*(m+m)v^2-ek1=(1/2)*(3+1)*1^2-8=-6j

wf=-umgs u= where s=3m

Slide the small slider slide up and down to see it as a whole, so s=2+1=3 meters).

When a small block stops, its kinetic energy is all converted into the kinetic energy of "a combination of a small block and a slider", and in this process, friction does negative work on the small block to reduce its kinetic energy (the total stroke, the magnitude of the work is the kinetic energy loss of the small block, according to which the frictional force can be obtained.

The first question: knowing the initial kinetic energy of a small block, you can find its initial velocity, and then according to the conservation of momentum, you can get the answer to the first question;

Question 2: Subtracting the total kinetic energy of the last two objects from the initial kinetic energy of the small block is the mechanical energy lost by the two objects rubbing against each other. Then the frictional force can be found.

bc If the direction of the electric field is horizontal, then the net force of gravity, electric field force, and rope tension force experienced by the ball is 0.

F Electro MG Tan 60

fElectricity q*e

The electric field strength is E mg*tan 60 q = root number 3 mg q at point b, the resultant force f of gravity and electric field force f is 2 mg and the angle of 30 degrees with the horizontal line, and the downward pull force t and f are along the vertical direction of the rope at an angle of 30 degrees to the horizontal line upward, by the triangle rule t = 3mg = qe

Regarding gravity and electric field force, it can be solved by the kinetic energy theorem, that is, W weight - W electricity = 0, that is, mgrsin60° = eqr (1-cos60°) to get eq = mg * root number 3, a is wrong, b is correct.

If point b is stationary, then there is sin60°=mg f, so f=2 3*root number 3mg, c is wrong.

Or cos60°=eq f, f=correct.

According to the symmetry, when the ball is at the midpoint of AB, the tangent direction resultant force is zero, and the angle between the thin line and the horizontal direction is exactly 30°, according to the trigonometric relationship, it can be obtained: QESIN 30° MGCOS 30°, simplification shows that option A is wrong, B is correct; When the velocity of the ball reaches point B, the resultant force along the direction of the thin line is zero, and the force analysis of the ball shows that FT QESIN 30° MGCOS 30°, and the simplification can know FT mg, and the options c and d are wrong

When t=0, the voltage across the capacitor is 0V, and when t, the voltage across the capacitor is 10V, so the function of the voltage across the capacitor with time is:

v(t)=v( )v(0)-v( )e (-t) where =rc = , so there is v(t) =10 - 10e (-t

When t=5 10 (-5), v = , q=cv= sorry, the power shouldn't be there... Even if the voltage at both ends of the capacitor is 10V, the charge of the capacitor is only 10 (-6)f 10V = 10 (-5)C....

If we follow the energy conservation column, 1 2 k x0 2 = 1 2 m + 3 m) v 2 +

There should also be a 1 2 k x 2 on the right, because the spring is not naturally elongated.

But how much is this x I don't know.

So it can't be solved this way.

The three objects are relatively stationary and have no sliding friction with each other.

A is constant, the force is balanced, and the friction force of A by b is f=0

b constant velocity, force balance, f-a to b friction f-c to b friction f'=f-f'=0,f'To the left, greater than or equal to f

c constant velocity, force balance, ground to c friction f"-b vs. c friction f'=0,f"=f'=f, the direction is to the left.

Summary: A is not subject to friction.

b is subjected to a frictional force to the left in the F direction of magnitude and so on.

c The upper surface is subjected to a rightward friction force in the f-direction of magnitude, and the lower surface is subjected to a left-facing friction force in the f-direction of magnitude.

335 3600 (60 1000) = 20 km h if viewed separately.

335 60 is given for the velocity of m s.

Then you need to convert m s to km h, so you need to multiply it again, and for the sake of calculation, it is usually written as 335 3600 (60 1000), which is a combination of many problems in physics, which is easy to simplify.

There is indeed a problem with parsing, it should be 335 60 (1000)=20 km h. Multiply 60 by 60 is 60 minutes per hour, and dividing by 1000 is 1000 meters per km.

Since it is a pipe, the velocity may be zero at the highest point, i.e. there is a support force.

The point is to have a proper understanding of the movement.

Brother,There's a problem with this question.,V's unit time m s,At least it's also the root number under the gr.,You check it out and then talk about it.,OK.

The usual echo practice problem is that the sound source is not moving, and the sound source of this question is moving.

Suppose the distance between the car and the high wall when the driver hears the echo is s, it is known that v car = 12m s, v sound = 340m s, t =

The car continues to move forward after honking, the driving time of the car and the time of sound propagation are, the distance traveled by the car is S car, then S car = V car T, and the distance of sound propagation is S sound = V sound T, according to the meaning of the question (draw a simple diagram by yourself).

S sound - S car = 2s, that is, V sound t - V car T = 2s, v car = 12m s, V sound = 340m s, t = substituted into the above equation to get the solution.

s=410m