Speed Problem, Physics Ace, Physics Problem Ace Fast .

Only the ideal state does not consider friction, and in this practical problem, friction must be considered!

When the initial velocity is 6 meters and seconds, the final velocity is still 6, according to the energy balance, there is no loss of kinetic energy, and the potential energy must be lost, set to p1And p1 is equal to the negative work done to him by friction during the motion of the object!

When the initial velocity is 5 meters and seconds, because the surface is concave, the centrifugal force must be considered, because the initial velocity becomes smaller, the centrifugal force becomes smaller, the pressure on the surface decreases, the friction between the object and the surface decreases, the displacement of the object before and after the two times is the same, so the negative work done by the friction force decreases, and the change of the potential energy of the object is also p1, and the change of the available kinetic energy is smaller than the first time, so the velocity is greater than 5!(It's okay to think about air resistance as well, because air resistance is proportional to the square of the object's velocity!) Since the impact is small, it can be ignored!

My end! Thank you!

The answer is greater than 5 meters in seconds.

At point A, the object has kinetic energy and potential energy, and at point B, it has kinetic energy. It is subjected to friction in the process of going from A to B. The work done by friction is equal to potential energy. But the work done by friction is related to the velocity, and the greater the velocity, the greater the work.

When the muzzle velocity is 6, the work done by friction is equal to the potential energy. Then when the muzzle velocity is 5, the work done by friction will be less than the potential energy.

So when the muzzle velocity is 5, it will end up being greater than 5

Or 5, there is a question to know that the horizontal height of a and b is the same, and the surface of the area is frictionless, so according to the principle of conservation of energy, the kinetic energy remains unchanged, and the velocity is still 5

1) >5ms

Consider that the start and end points are not equal in height. When the pressure of the object on the surface is large, so the friction is large, when it becomes 5m s, the friction force becomes smaller, and the work consumed by the friction force is also reduced, leaving a part of the potential energy converted into kinetic energy, so the velocity is greater than 5m s

2) Equal to 5m s

If the starting point and the end point are considered to be equal in height, it means that the surface is smooth and the friction force does not work. is still 5m s

Still 5, no energy loss.

c……Need to explain?

The first two must be fine.

c d depends on the density, and the control variable is just fine.

That's ......You see, let's start with A

When there is a very small aluminum ball in the hand and a huge copper ball, both the mass and volume of the aluminum ball are smaller than the copper ball.

So, b pass

Then look at C and D, aluminum is less dense than copper, right? Therefore, when the mass is the same, the volume of aluminum should be larger, but how can there be a situation where the mass is larger than copper, but the volume is not smaller than copper? So, c is wrong, and in the same way, it can be explained that d, so d is right.

Well, the answer is c

A and B are both possible.

The mass is the same, the aluminum is large, and the volume is impossible.

s=at^2/2

a=1m/s^2

v=at=4m/s

The speed of the car is 4m s when the engine is turned off

a1=v/t1=2m/s^2

f=ma1=4000n

f-f=ma

f=6000n

Under the action of constant traction, a car starts to move in a straight line from stationary, so do a uniform acceleration to close the engine, and decelerate evenly, and the friction force is the resultant force.

The homogenization acceleration stage was analyzed for f-f=ma1

The uniform deceleration phase was analyzed for f=mA2

From the meaning of the title, it can be seen that s1=

Agree to count as A2

Regardless of the resistance of the water, momentum is conserved.

When the frog jumps off the leaf armor for the second time, the leaf armor receives three jump momentums, and the first time mv1=mv 2, v1=v 2 jumps away.

The second mv2=mv1+mv 2=2mv 2 v2=v jump.

3rd mv3=mv2+mv 2=3mv 2 v3=3v 2 jump.

The frog jumps off the leaf carapace for the second time at a speed of 3v 2

1: mA*vA1=mfrog*v=ft, vA1=V2

2: ft = m frog v = (m A + m frog) v total.

3: (M A + M Frog) v total = M A v A 3-M frog V, let the direction of A be positive, so V A = 3 2V

1 All (1).The distance from the 108 to the 113 milestone is 5km and the car takes 3 minutes and 30 seconds. So the speed of the car is about 5000 (3*60+30)=

2).The distance the car has traveled in 1 minute and 40 seconds. So the speed of the car is 2400 100 = 24m s=

3).The results are in good agreement with those in (1) and (2).

4）.In cooperation with classmates, one person looks at the watch (the current electronic watch, or mobile phone has a stopwatch function, which can improve the measurement accuracy), and the other person looks at the boundary pillar, and agrees to give a signal immediately every 1 kilometer of boundary marker. In this way, 10 kilometers are measured, i.e. ten sets of data are measured, and then the average is taken, so that the data obtained is more accurate.

One. 1) Volume = 20 * 10 * 6 = 1200 cm 3 Mass = Density = 1800 1200 =

2) Gravity = mg =

Base area s1=20*10= p1=900pas2=20*6= = p2=1500pa

s3=10*6= = p3=3000pa

3) The maximum storage is 4500 900 = 5 pieces.

Two. A 5 laps B 4 laps The time is the same then v A: v B = s A: s B = 5: 4 A speed 5000 24 * 60 1000 =

B speed = B time = 5km 10 = i.e. 30 minutes.

Yijia arrived at the finish line 6 minutes late.

3 g takes 10 n per kilogram to indicate that it is calculated according to the standard air pressure.

Air density. The air thickness is taken as 3000 meters.

Then the volume of air = (6370,000+3000,000) 3)-6370,000 3)4 3

10^18 m^3

m 3 mass = kg