Who s going to do this physics problem! Who will do this physics problem? Help write the answers!

You're talking about uniform acceleration

Let the distance of OA be x and the acceleration be a

Movement of the OA segment: S(OA) = X, T(OA) = (2S A) Movement of the OB segment: S(OB) = X + I, T(OB) = (2S A) Movement of the OC segment:

S(oc)=x+i+d,t(oc)=(2s a) Time through ab: t(ab)=t(ob)-t(oa) Time through bc: t(bc)=t(oc)-t(ob) Because t(ab)=t(bc), the joint equation obtains:

The distance of 2(x+i) oa: x=(9i 2+d 2-6id) 8(d-i)

It is known that the object moves at a uniform speed through the AB segment and the BC segment in equal time.

The distance between AB is L, and the distance between BC is D

So: ab=bc=1

However, it has nothing to do with OA.

No way to find: find the distance between o and a.

I can't seem to do it, maybe I'm just a shallow learner.

Take a step late and let someone else solve it.

Let's try a system of simultaneous ternary equations.

Let the acceleration a and the time t

v=att=1.

vt+1 2at 2=at-1 2a=10 at the last second, v1=v-a(t-1)=a

According to the formula v2-v0 2=2ax

a^2-0^2=4a

get a=4t=3s

Density: (p alcohol = cubic centimeters).

Mass = P alcohol * V = cubic centimeter * 500ml = 400g

"45" means that 100ml of liquor contains 45ml of alcohol and 55ml of water, so the quality of this 100ml of liquor is the sum of the two.

The volume of alcohol * the density of alcohol + the volume of water * the density of water and the mass of water divided by the volume (one hundred milliliters) can get the density of the liquor, the density of the liquor multiplied by the volume of the liquor five hundred milliliters can get the mass of the liquor, and the calculation part will be processed by itself.

According to the conditions, the volume of alcohol in the whole bottle of liquor v1 = 500ml 45% = 225ml, and the volume of water v2 = 500ml-225ml = 275ml

We know the density of water p water = grams of cubic centimeters, and we tell p alcohol = cubic centimeters in the title, we can find the quality of water in the wine and the quality of the alcohol, and the sum of the two is the quality of the bottle. With the formula (m=pv).

Find out the quality, and then use the mass 500ml is the density of the reminder, do this kind of problem to pay attention to the unit.

vWater = 500*

v alcohol = 500-275 = 225ml

Mass m = v water * 1 + v alcohol * = 455g density p = m v = 455 500 = cubic centimeters.

Will the object you hit move with the car?

If not: Solution: ft=mv=4000*50 Because there is no time in the condition, there is no solution. ）

If so, it is not possible to find the mass of the object because there is no mass in the condition.

The impact force of the car on the object is related to the length of the impact time. There is no impact time and it cannot be solved. It's like a person hitting a sponge, and the force on hitting a concrete floor is definitely not the same.

The first question is the kinetic energy theorem. The second question is kinematics. It's very simple to do it directly. Won't ask again.

I'm sorry, I don't understand the problem, but you should be able to solve it mathematically.

Because the displacement of the last 1 second is 2 meters, the average velocity is 2 m s, and the velocity at the end of the last 1 second is 0, which is the acceleration of -4 m s according to the proportion

And the average velocity of the first second is 10m s There are many formulas that can be used, use the simplest one to average the speed: the average velocity of the first second is 10, the second is 6, and the third second is 2 displacement: the first second is 10, the second is 6, and the third second is 2

Total displacement = 10 + 6 + 2 = 18

Uniform deceleration linear motion is a reduction in the displacement unit time, the displacement in the last second is 2 meters, and the displacement in the next second is 0 meters, so the displacement per second is reduced by 2 meters, so the result is 10 + 8 + 6 + 4 + 2 = 30 meters.

In the last 1 s, it can be seen as a reverse acceleration with the initial velocity of 0, then x=1 2at is substituted into the value, note that because it is a deceleration, a=-4m s and then go back and look at the first second.

x=v, t+1, 2at, v=12m, s, 2ax=v-end, -v-end, v-end, v-end, v-end, v-end, v--v--v-v-

The solution is x=18m

The displacement in the last second is 2 meters, i.e. s = 1 2 x a x t22 = 0 - 1 2 x a x 1 2

a=-4m s2 (the negative sign indicates that the direction of motion is opposite to the direction of acceleration) is 10 meters in 1 second, i.e. s=vo x t - 1 2 x a x t2

10=vo x 1 - 1/2 x 4 x 1^2vo=12m/s

Total displacement: s = (0 - vo 2) (2 x a) = -12 2 ( 2 x -4) = 18m

s1=v0*t1+1/2a*t1^2

a=合=fcos37-f=ma

f=2n.After removing f, the resultant external force is equal to the friction Xiaokai force = 2n, and the acceleration is cautious a'=f/m=1.

The velocity of pure land when f is removed v=v0+at1=2m s

vt^2-v0^2=2as

s=2m

The acceleration obtained from s1=v0*t1+1 2a*t1 2 is obtained by a=combined=fcos37-f=ma, and the frictional force f=2n. is obtainedAfter removing f, it is judged that the external force is equal to the friction force = 2n, and the excavation velocity a is added'=f/m=1.The velocity at which f is removed is v=v0+at1=2m s

The displacement = 2m is obtained from vt 2-v0 2 = 2as

Let the fall time t and the degree of gravitational acceleration be g

h=1/2 g t^2

32=1/2 g t^2

32-14=1/2 g(t-1)^2

t=4s g=4m/s^2

Look at the work formula of accelerating motion evenly, v 2 = 2as

Bring it yourself.

Let the fall time t and the acceleration due to gravity be g

Last-second s=-formula.

Total distance s = two formulas.

One-two joint solution.

The solution yields t=4s, g=4m, s2