A 20 point reward for sophomore physics and a high school physics question for a high score

Think of wires of different thicknesses as being tied together with the same thin wires with a turn number of n.

Let the number of turns of the coil be n, the length of each side be l, the density be , the resistivity be ', and the cross-sectional area of the wire is s.

then the mass m= ·4nls

The velocity of the two coils when they fall freely into the field of the magnetic field is the same (kinetic energy theorem), according to the time when entering the magnetic field.

to determine their motion.

Because e=nblv, i=e r=(nblv) ( '4nl s).

And because ma=mg-nbil

So a=g-nbil m

Substituting i and m into it, we can finally get a=g-b v 16 '

Obviously, a is a fixed value, because the initial velocity and acceleration of the wireframe after entering the magnetic field are the same, the landing time is the same.

The heat generated q=eδtr=n δ δt·r=(nbl) δt·r

Because m= ·4nls , q=b l ·m 16 't

The heat generated by the motion generated in the electromagnetic field is independent of the number of turns n, so the heat is the same, a is correct.

Select du=blv

p=u*u/r

However, due to the small resistance of the thick wire and the large current, the thick wire can be regarded as n identical thin wires tied together, so it can be seen that the heat generation of the thick coil is large, and the two coils fall to the ground at the same time.

Let me add: let the resistivity be a, the density be b, and s be the cross-sectional area, then r=al s and m=bv; v=4sl;

f=mg-b^2 l^2v/r;Acceleration = f m = g - b 2v 4b is a fixed value.

Therefore, it landed at the same time.

p=u*u r=b 2 s v 2 a shows that a thick coil generates a lot of heat, so there is no correct answer.

The object is subjected to 4 forces, frictional force f

n,n=mg-f

sin37°So.

Friction to do work should be.

w=fs=μ(mg-f

sin37°)*s.。

sin37° is the component of the tensile force in the vertical direction. Reduced stress as a result. ok？

Doesn't the diagram seem to be of much use? Answer: AC

I don't know how to ask questions. About the estimation: The ohmic file does not need to be estimated, because the scale of the ohm file is uneven, and the estimated reading value cannot be estimated, as long as the nearest value is read (this requires a shift comparison).

Whether or not it is estimated depends on the minimum index of the dial. For example:

Suppose the ammeter, the smallest division, then you can't estimate it, and if you read it, then others can't tell if the number you read is an accurate value or an estimated reading, because the number of digits is the same.

If the minimum scale of a dial is, then you have to estimate, for example, so that others will know that the last digit is the estimator.

First, set vb=xm s and list the following equations:

Rise time t1 = vb g

Ascending distance s1 = vb*t1-a*t1 2

Fall from the highest point to the surface of the water (3+s1)=a*t2 2 Supplementary equation t1+t2=3s

Simultaneous four equations vb=14m s t1= s1=velocity of falling into the water, according to equation 2*a*(3+s1)=v 2.

That's the answer I made for you!

(1)x=v0*t+(1/2)at^2=4+(1/2)*3*4=10m

a=(8m s-2m s) 2s=3m s 2(2) is obtained by Newton's second law f = ma: mgsin -f=ma, i.e.: 375-f=225

So: f=150n

The direction is upwards along the inclined plane.

The force analysis shows that mgcos30°-n=0;mgsin30°-nu-f air resistance=mA; From the data of the question, we know that a=(8-2) 2=3m s from vt -v0 =2as, s=10m, total resistance, f=nu+f, air resistance=150n

According to the inscription, it can be seen that in the process of moving a small block from point m to point n, the friction force (rough horizontal plane) and the point charge q exert on the block (electric field force) in the horizontal direction opposite to the direction of motion, and the friction force is opposite to the electric field force.

Since the block is moved from rest by the force of the electric field, the direction of the electric field force is the same as the direction of motion, and the direction of friction is opposite.

The tangent direction of each point of the electric field line is consistent with the direction of the field strength at that point. The electric field line can only describe the direction of the electric field and qualitatively describe the strength of the electric field, which determines the direction of the electric field force and the direction of acceleration (Newton's second law) of the charged particles at each point on the electric field line, and the electric field line is not necessarily the trajectory of the charged particles in the electric field.

LZ: Not very complete.

Small Blocks with Positive or Negative Bands. is the key to judgment.

One. It's with the right ... is to the left. Positive: The direction of motion is the same as the direction of the electric field.

Two. Negative means to the right. Negative: The direction of motion is opposite to the direction of the electric field.

The direction of the field strength can be determined by looking at the electric field lines, but the force can also be determined according to the motion state of the object: in this problem, "no initial velocity release", moving to the left under the action of the electric field force, the force is of course to the left.

Since the block has no initial velocity, it is only subjected to electric field forces in the horizontal direction.

It moves to the left, so it can only be subjected to the electric field force to the left.

The direction of the electric field force is not necessarily determined by the direction of the electric field lines.

Sometimes it also depends on the electrical properties of the charge

The speed of the river has no effect on the boat.

Mr. Hoho has talked about this question.

The puppy runs for the same time as A and B walk.

That is, 12 (5+3)=

So the puppy's distance is 6*

The puppy finally stops at the place where A and B meet.

That is, 5* away from the departure place of A.

So the displacement of the puppy is.

Because the time for the three animals must be 12 =

So the puppy also walked the way.

1.Rope tension: f=m(a+g)=5kg x ( 51n People are rubbed by the ground: f=fsin37*=

2.The upward pull of the rope on the person: f. =fcos37*= human gravity: g=mg=70kg x 10m s*=700n

g=f。+n The pressure of a person on the ground n=