Help, prawns, a math or physics problem .

There is a problem with the question "it takes 40 minutes longer to return the result than it does when you go", so the original question is not solved.

Proof] Let the distance at the time of departure be s.

The time it takes to walk is T1 and the distance to get there is T2.

t1=s 2 v walking

t2=s 2 v ride

then go to the time. v go time average = s (t1 + t2).

Substitute , then.

v Departure Average = (2V Walk * V Ride) (V Walk + V Ride) Let the distance at return be s.

The distance on foot on the return is S1 and the distance by car is S2.

s1=v walk *t 2

s2=v ride * t 2

then when returning. v returns the average = (s1 + s2) t

Substitute , then.

v returns average = (v walk + v ride) 2

So in summary, if v walk=v1, v ride=v2, there is.

v returns the average v when going to the average = (v1+v2) 2 2v1*v2 (v1+v2).

v1+v2）^2－4v1v2]/（2v1*v2）（v1－v2）^2/（2v1*v2）

0 (easy to know v1 0, v2 0).

Therefore, v returns the average v when going to the average, and because the distance at the time of going and returning is the same, v returns the average v the average when going to the time, then the time taken to go should be more than the time taken to return.

However, the title mentions that "it takes 40 minutes more to return than it does when you go" is not valid, so the original question is unsolved.

Let's draw a picture, it's not very difficult, but the steps are more complicated, and you have to set an unknown number first.

The gravity of hollow iron is 2000N, take g = 10, then the mass of hollow iron is 200kg.

The density of iron is taken, then the volume of iron is 200 7860 = cubic meters.

So the hollow part is cubic meters.

Solution: (1) t 1 = 30 min = ......1 point) ......2 points).

…2 points) 2) The shortest time t 3 ......1 point) ......

The time to the exit is 11:30

The maximum time t 4 ......1 point) ......

The time to the exit is 13:00

Xiao Ming told his dad that the time slot was 11:30 13:00

1) 18 (30 60) = 36 The speed of the car is 36km h

18 40) 60 = 27 It takes 27 minutes to get from the sign to the bridge.

Xiao Ming told his father that it was between 1 hour and 30 minutes and 3 hours.

1.30min = 18 18 40 = 180 120 = 3 hours and 1 and a half hours from 10 o'clock onwards, respectively. The resulting times are 11:

30 and 13: 00 so. The time period Xiao Ming said was 11 a.m

30 - 1 p.m.

1) v = 18 t = 18 40 = minutes (2) t minimum = 180 120 =, t maximum = 18 60 = 3h, so you should arrive at the exit of the expressway between 11:30 and 13:00 to avoid violations.

First, the mass of the object is: m=m2-m3=800g-200g=600g;

The volume of the object (equal to the volume of water discharged by the object) is:

v = v row = m drain water = (m1-m3) water = (500g-200g) ;

The density of the object is: =m v=600g 300cm=.

Second, the system of equations: m copper + m aluminum = m total =;

m copper copper + m aluminum aluminum = v total = 3cm . The density value is used,. Calculations are simpler).

The solution yields: m copper = m aluminum =

(1) V matter = 500-200 p water = 300cm 3m matter = 800-200 = 600g

P = 600 300 = 2g cm3

2) P copper * v copper + p aluminum * aluminum =

v copper + v aluminum = 3

The landlord didn't give the density of copper and aluminum, so I can only do this, bring it in and calculate it myself.

1.From the change of the mass of the cup after taking out, it can be seen that the weight of the object is 800-200=600g; Excluded water 500-200 = 300g. Archimedes' brother tells us that the volume of the object is the same as the volume excluding water (the conditions are insufficient, pretending to be sunk to the bottom), and the density can be obtained from 1000kg m3*600 300=2000kg m3

2.The total density, consisting of m copper + m aluminum = m and m copper p copper + m aluminum p aluminum = v easy to obtain m copper = m * (, m aluminum =

1.Landlord, you didn't say anything about the floating of this object.

If it is completely submerged, then v = v drainage.

m water = 500g - 200g = 300g, v drainage = m density = 300cm

m = 800g-200g = 600g

So V object = v drainage = 300cm

600g Object density = 300cm

That is, the density of the object = 2g cm

If it stops at point A, it will continue to slide, so it will not stop at point A, and there is acceleration at point A and point B with the same force. Gravitational potential energy is converted into elastic potential energy and kinetic energy as well as frictional heat generation. The kinetic energy is zero when the elastic potential energy is maximum, and the elastic potential energy is not zero when the kinetic energy is maximum.

The preliminary judgment is that the kinetic energy is less than the elastic potential energy. How to consider the loss of friction? The maximum kinetic energy and part of the gravitational potential energy are converted into elastic potential energy and frictional work.

As part of the gravitational potential energy must be greater than the frictional work. So the kinetic energy of the force must be less than the potential energy of the elasticity. Why must the gravitational potential energy at this stage be greater than the work done by this order of friction?

The friction is constant, the gravity is also constant, or back to point b. If the work done by friction is greater than the gravitational potential energy at that stage, then he should have stopped a long time ago, and would not have slid down, or even slid up. So the frictional workmanship is less than the gravitational potential energy, so the elastic potential energy is greater than the maximum kinetic energy.

<> don't know why, ** can't be inserted.

With regard to option C, it was related to . When the particle drops to the height h, the gravitational potential energy decreases mgh and the kinetic energy rises mgh*(tan ) 2It depends on the size of 45 degrees.