Ask an interesting question about physical motion

The Earth rotates at the same time as it rotates and maintains a certain angular velocity, so leap years are generated, which means that the rotation of the Earth cannot be considered alone. (The simplest example is that the day-night cycle in the Arctic and Antarctic Circles is not due to the rotation of the Earth, but to the revolution of the Earth).

Therefore, in an ideal situation, no matter how the car moves, as long as the angle between it and the earth and the sun is certain (that is, the triangle formed by the three points of the car, the center of the earth and the sun only rotates around the one point of the sun, and does not rotate around any other point or side), he will never have a day and night cycle!

That is to say, only the speed of the car should change with the rotation of the earth, and the direction of the car should also change with it, only the speed should be the same, and it is also possible to see the change of day and night.

There are two scenarios.

1 The car travels in the same direction as the Earth's rotation, so the person sitting in the car will not have a day and night cycle.

2 Otherwise, there will be an alternation of day and night.

It's just the opposite of what the second floor says:

1.The direction of the car is the same as that of the Earth's rotation, with an alternation of day and night.

2.Conversely, there is no alternation of day and night.

First of all, it is assumed that the movement of the car is on the equator line, and if it moves along the meridian line, there must be an alternation of day and night.

1).In the opposite direction of rotation, there is a day-night alternation, and the driver feels a cycle of 12 hours.

2).As with the direction of rotation, there is also a day-night alternation, and the driver perceives the cycle as one year. Because although the driver feels that there is no rotation anymore, do not forget that the earth still rotates.

If you have the conditions, you can take a globe and try it, determine a point, and there is no rotation, you will find that this point shines on the sun for half a year, but not for half a year.

Absolutely.

For example, trying to measure the depth of seawater with a wire is often not possible, because as the depth of the sea increases, the wire lengthens, and when the maximum tension of the wire is exceeded, the wire is broken. You may think, the maximum tensile force of thin steel wire is small, why not use thick steel wire? If you think about it, no, the wire is thicker and the gravity increases, so it is impossible to measure the seafloor at a depth of about a kilometer.

The above analysis is qualitative, and if it is analyzed quantitatively, this problem is still a bit difficult.

No, the object hanging on the rope breaks because the rope is stressed more than it can withstand at this node. And the gravity of the whole rope does not act on one point, so it will not break.

b right. The velocity of the particle is.

v dr dt 5* t 2 i 0 j that is, the particle is doing linear motion, and the velocity is v0 0 at t 0;At t 2 seconds, the speed is v1 5*2 2 20 m s

The work done by the kinetic energy theorem is obtained by the resultant force sought.

W (m*v1 2 2) (m*v0 2 2) m*v1 2 2 2 20 joules.

v=dr/dt=5t^2i

When t=0, v0=0, and t=2s, vt=20i

According to the functional principle, w=ek-ek0=m(vt 2) 2-0= j choose answer b

The mass of the carpet is evenly distributed.

then the part of the center of mass is pulled up at its geometric center.

From this, it can be seen that the displacement of the centroid is always 1 2 of the displacement of the moving end, and its velocity is 1 2 of the velocity of the moving end

When seeking force, the process involves the balance of an object in general.

Specialist help is available.

Since the velocity of the bullet is very large, the drag force is also very large (the drag force is proportional to the nth power of the velocity), then the work done by the drag force will be very large. When the bullet returns to its original altitude, the displacement is equal to zero, the kinetic energy of the change in gravity is zero, and the work done by the drag force (i.e., the loss of kinetic energy or mechanical energy) is very much. The lethality will be greatly reduced.

This refers to a purely physical explanation, which is bound to differ from the actual situation.

2) Is the lethality of a bullet measured only by its velocity?

It's dangerous, as you said, there's still a lot of momentum when the bullet falls, and injuries still happen, you can look at the news.

The first step is to calculate the time it takes for the wooden box to move when the ball falls.

According to the inscription, the wooden box is subjected to a net force before the small metal block falls.

f=9n-f=9n- mg=

According to s=12at2=12(fm)t2

Yields: 1=(1 2)*(1 2)t 2

Because the wooden box is 1m long, it is s=1, and the upper surface of the wooden box is smooth, so only the wooden box does the acceleration movement, so m=2kg).

The solution is t=2 seconds<3 seconds, so it is in line with the topic, and after the small metal block falls, the wooden box will continue to be subjected to constant force for 1 second.

In the second part, calculate the speed of the wooden box when the metal block falls.

v=at=(1/2)*2=1m/s

The third step is to calculate the distance that the wooden box moves after the small metal block falls, that is, the distance between the small metal block and the left edge of the wooden box after it is stationary.

In 1 second when the constant force continues, the net force experienced by the wooden box is 9-(, then the acceleration is 4 2 = 2m s 2

The travel distance is vt+(1 2)at 2=1*1+(1 2)*2*1=3m

And the final velocity is v+at=1+2=3m s

After the constant force is removed, the net force of the wooden box is only frictional force-(

Then the acceleration is -5 2=

The travel distance is vt+(1 2)at 2=3*t+(1 2)*(

The movement time is 3 (seconds (because it is decelerated from 3m s to stop, and substituted by the above equation).

s2 = the distance that the wooden box moves after the small metal block falls, that is, the distance between the small metal block and the left edge of the wooden box after it is stationary = 3+

mg-kv^2=ma=m(dv/dt)

t=0 v(t)=0

Solving differential equations is obtained, and the solution is not annoying, but the expression is too troublesome to knock here, you add 50 points to me, and I will write the answer.

Because it's a free-fall movement.

So v=gt

H = (1 2) gt quadratic.