After class exercises for Chapter 6 of Physics online and other answers !!

3: Increase. It is usually undetectable, because what we call "detection" is measured with corresponding instruments in classical physics, while the change of mass with velocity is non-classical physics, which is derived by formulas, so it is usually undetectable.

4: No, you can't. The force acts on an object, and the object does have some acceleration, but this is limited to classical physics. In non-classical physics, the law of the ox does not apply, so it cannot exceed the speed of light. Albert Einstein also introduced that no object can be faster than the speed of light.

5: Gets bigger. Becomes smaller (weightless).

Of course, my explanation doesn't necessarily apply to test-oriented education, because test-oriented education is often not the same as the truth (haha, off-topic).

3.Enlarge. Usually we can't detect it!

4.No! Because the energy of an object that reaches the speed of light is infinite!

5.Big! Constant motion is weightless, gravity is 0, of course small!

I don't understand, this is the theory of relativity, I won't take the college entrance examination, even the teacher can't explain it very clearly, it's too difficult, just know a little bit, the more you study this thing, the more confused it becomes.

Please ask if the 3 questions are written correctly, it feels a little wrong.

The third question is about the theory of relativity, what is the reason for studying this in the first year of high school, the mass should increase, the theory of relativity says that when the velocity is very large, the mass of the object will increase relatively, but this is only Einstein's assumption that has not been proved by experiments, so I don't know if it can be detected.

Very interesting question:

3.The mass becomes larger, but we can't detect it, and when v can't be compared to the speed of light, the mass of the object changes too little.

4.The velocity of any object cannot exceed the speed of light, because by f=ma, this great force acts on the object, the acceleration of the object is large at the beginning, but as the speed increases, especially when the speed is close to the speed of light, the mass of the object is surprisingly large, then the acceleration is very small, when the object is slightly equal to the speed of light, the mass of the object is infinite, and the acceleration is slightly 0, so the maximum speed can only reach the speed of light.

5.It is greater than the gravity on the earth, and if the speed is constant, it is in a state of complete weightlessness.

(1) Hearing 66 sounds, that is, walking 65 sections, so after walking x=65 25=1625m, 1 and a half minutes, that is, t=90s, so there is v=x t, v=1625 90=18m s.

2) Find the average velocity in the first 2s, there are x=1+3=4m, t=2s, v=4 2=2ms

To find the average velocity in the last 2s, there are x=5+7=12m, t=2s, v=12 2=6ms

To find the average velocity of all motion time, there is x=1+3+5+7=16m, t=4s, v=16 4=4m

Kai talks about h=10m s

s=v1*t1+v1*t2+

250(m)

v flat 1 = s t = 250 20 =

v Ping 2 = (S-V1*T1) T2 = 150 10 = 15 (m s) v=a*t1 = 2 * 5 = 10 (m s) branch base.

50 = t3 = 10 (Mengsun Jin S).

t=t1+t2+t3=5+120+10=135(s)

Pick c, the calculation of work is the force multiplied by the distance in the direction of the force w=fs1) The work done by the force is completely converted into the kinetic energy of the object.

2) In addition to the kinetic energy done by the force, there is also a part of the work done to overcome the frictional force, so the kinetic energy of the object in this case is less than (1).

3) In addition to the kinetic energy of the object, the work done by the force is also partially converted into the gravitational potential energy of the object, because the object is pushed upward along the inclined plane and the horizontal height is also increased, and the kinetic energy of the object is also smaller than (1).

Hope it helps!

f work w = fs, s is the displacement of the object along the f direction, so the answer is: c

Choose as much as c Detailed explanation: The amount of work done by force is the product of the force and the distance of the object in the direction of the force, and has nothing to do with the mass of the object and other quantities.

Selecting c can be solved using the formula w=fs.

The three works done are all fs because the work done by constant force is only related to the displacement and angle in the direction of the force and has nothing to do with the others. So choose answer C

1) The horizontal thrust f pushes an object with a mass m to advance on a smooth horizontal plane s, and the work done by force f is w = fs

2) The horizontal thrust f pushes an object with a mass of 2m along the horizontal friction factor of s and advances in front of s, and the work done by force f is w=fs

3) The inclination angle of the inclined plane is , the thrust f parallel to the inclined plane, and an object with a mass of 2m is pushed upwards along the smooth inclined plane for s, and the work done by the force f is w = fs

So choose C as much as (because w=fs).

h=vt+1/2gt^2

where v=3 h=60 solution t

After solving t, you can enter vt=v+gt and solve vt.

1) From x=v0t+a=2(x-v0t) t 2, substitute the data x1=24m, x2=64m, t1=4s, t2=8s, get a=2(24-4v0) 16 m s 2=2(64-8v0) 64 m s 2, and get v0=4m s, a=1m s 2

2) Let the acceleration be a, the time is t, and the velocity is v, then when t=1, v1=a; t=3, v2=3a; t=6, v3=6a. So the average velocity of the object when passing through time is a a+a a + 3a 2, so the ratio of the distance passed is 1:8:

The 100-meter acceleration time is T1, the constant speed running time is T2, the 200-meter acceleration time is T3, and the constant speed running time is T4.

v=at1100=at1^2/2+vt2

t1+t2=

200=at2^2/2+

t3+t4=

t1=t3 to solve the simultaneous equation.

I will first analyze the method for you in this question, first of all, it must be conserved with mechanical energy. But the system of A and B should be analyzed, because if you analyze A or B balls separately, A and B are respectively affected by the spring force and the mechanical energy is not conserved. So it should be analyzed on the whole.

Now let's start analyzing the physical process, since the elastic force of the spring does not change abruptly after withdrawal, the A ball will be subjected to a force that is very opposite to M2G, and the downward acceleration motion will be made, and the B ball will still be in equilibrium at this moment. Before removing f, the tensile length of the spring x1=m2g k can be obtained from Hooke's law, and the elongation of the spring after landing is equal to m1g, x2=m1 k, and according to the conservation of mechanical energy of the system e1=e2, the original length of the spring is l to obtain m1g(l+x1)+m2gh=m1g(l-x2)+m2g*0;Thus k, according to ep=1 2kx 2 (x is the type variable) where x=x2, ep can be obtained

I hope it helps, if you don't understand, ask me.

At the moment when F is removed, since the spring force does not change abruptly, ball A will be subjected to a force that is opposite to that of F, and it will accelerate downward, and ball B will remain in equilibrium at this moment. Before removing f, Hooke's law can be used to find the tensile length of the spring x=m2g k. Let the original length of the spring l, and take the ground as the zero potential energy surface.

The B ball just touches the ground and the A ball falls at a height of (h+x), the decreasing gravitational potential energy is m1g(h+x), and the elastic potential energy that the spring had before is 1 2kx. By the conservation of mechanical energy, when the elastic potential energy of the spring is maximized again, that is, the mechanical energy of the ball A is all converted into elastic potential energy, for.