Urgent high school physics questions hope to get the answer today about speed

v=15m/s 30cm/s2=

The uniform and subteriotic motion can be regarded as the additive motion with an initial velocity of 0, according to v=at==>t=va=50s

The time when impulse 0 accelerates to 15m s is s=15

The displacement of uniform meiotic motion s=1 2at2=750m, the displacement of the acceleration from impulse 0 to 15m2, s=1 2at2=450m, the displacement of the delay s=15*80-1 2(1200)=600m, the time of the delay, s=600, 15=40s

If it's 100s, then the question is wrong, because it should stop for 1 minute, not 1s, and look at the train and stop it for 1s

v0=54km/h=15m/s

It is obtained by vt 2-v0 2=2as.

15 2 = 2 * s = 225 m v flat =

Brake delay time = 225 (

Parking delay time = 1s

Start delay time = braking delay time = 30s

Since the time delayed by the pause = 30 + 1 + 30 = 61s is not 100 seconds, it is 61 seconds

54 km/h = 15 m/s.

15 seconds. 15 seconds.

50 + 30 = 80 seconds.

225 seconds.

The next question is to find the distance traveled by the train.

Everything is OK!

Let the total displacement be x, then the first 1 and a half distance of the glide orange game is x 2

x=vt1t1=x/12

In the same way, t2 = x 20

The average value of the velocity is x (t1 + t2) = x (x 12 + x 20) =

Let the total distance be s, the first half of s s 2 takes s 12 seconds, the second half of s s 2 takes s 20 seconds, and the total answer time is 2s Qing Zhengbei 15 seconds, so the average speed is Cong Jins

When the Lub and liquid thoughts are each half, the formula is quickly calculated and the buried trace is calculated.

v (average speed) = 2 * v1 * v2 (v1 + v2) = 2x6x10 16 =

Let the total distance be 2s, and the first half of the distance will take the following time

t1=s Zen burial v1=s 6

The second half of the journey is called by the time and shirt for:

t2=s/v2=s/10

So the total travel time is:

t=t1+t2=s 6+s 10=4s 15 so the average velocity v is:

v=2s/t=(2s)/(4s/15)=

Let the uphill speed be v1 and the downhill speed be v2 then:

v1=2 3v2 t1= 2 5s v1+3 5s v2 when going to school

3 5s v2+3 5s v2 when going home t2 = 3 5s v1+2 5s v2 = 9 10s v2+2 5s v2 up and down divide to get the relationship between t1 and t2, i.e., t1 t2 = 12 13, so t2=13 12t1 t2=39min

Let the downhill speed be v and the distance is s

The uphill speed is 2v 3

Going to school 2 5 Uphill 3 5 Downhill.

That is, (2s 5) (2v 3) + (3s 5) v=36min, simplified to 6s 5v=36min, and obtained sv=30min after school, the original uphill road at school becomes downhill, and the original downhill road is programmed uphill, that is, 3 5 uphill and 2 5 downhill.

Then the time is: (3s 5) (2v 3) + (2s 5) v=13s 10v=(13 10)*30=39min

+ The time it takes for the ball to fall from a high place and hit the table is:

t1=√（2h/g）=√39/49)

v1=(7/9)*t1*g

Due to its upward muzzle velocity, the ball moves upwards vertically upwards and moves in free fall after reaching the highest point, and the time taken for both phases is the same

t2=2*(v1 g)=2*(7 9)*t1v2=(7 9)v1=(7 9) *t1*gSo: t3=2*(v2 g)=2*(7 9) *t1v3=(7 9)v2=(7 9) *t1*gtn+1=2*(7 9) n*t1 (n>0) so: t total = t1+t2+t3+....tn+1

2*[(7/9)*t1+ (7/9)²*t1+（7/9)³*t1+…(7/9)∧n*t1]+t1

2*(9 2)[(7 9)-(7 9) (n+1)]*t1+t1When n, (7 9) (n+1) 0

t total = 2 * (9 2) * (7 9) * t1 + t1 = 8t1 Answer: The time taken by the ball from the time it starts falling to the time it stops moving.

I don't know if it's wrong, but if it's wrong, please forgive me. I have done a very similar problem before, but the height of the ball is the same, other conditions are the same, at this time only t1 changes, and the final answer is 8s

v2=2gh

Find an equal proportional series, when n is infinite, it will do.

Is there any air resistance?

If not, find a proportional series, n take 21

h=1/2gt^2

t=1sv=at

v = n = the first term is 1, the common ratio is 7 9 in the proportional series, and then find the sum of the first 21 terms in the proportional series.

This sum minus 1, multiply by 2, add 1, and it should be fine (maybe this number of terms is a bit problematic, I can only do so much, haha.)

Let the time of the object's motion be t, then the displacement of the first 1 3 is 1 3*t*v1, and the displacement of the last 2 3 is 2 3*t*v2

The displacement of the two segments is added and divided by time t to give the average velocity:

1/3*t*v1）+（2/3*t*v2）】÷t = 1/3v1+2/3v2

Set the time to t

The average speed is the total distance divided by the total time, i.e.: (v1*t*1 3+v2*t*2 3) t=v1 3+2v2 3

It's easy! Let the total time be 3t, then v1*t+2t*v2=s total distance;

Then use the total distance s (3t) which is the average number of degrees.

The beginning of the object is subjected to four forces:

friction force f, along the direction of the inclined plane, up and down unknown;

The inclined plane supports the force n, and the vertical inclined plane is upward;

gravity g, straight downward;

f1=2n, down along the inclined plane;

f2 = 10n, up along the inclined plane.

If the object is at rest, the frictional force experienced by the object is equal to the other combined external force except the frictional force, and the opposite is true.

In the direction along the inclined plane:

The gravitational component GSIN37°=20*, is downward along the inclined plane.

gsin37°+f1=12+2=14n ＞ f2=10n

The sum of the external forces other than the frictional force along the inclined plane is 14 - 10 = 4n, and the downward direction is along the inclined plane.

So the friction force f=4n, along the inclined plane.

Synthesis: f = GSIN37°+F1 - F2 = 12+2-10 =14N].

If f1 is removed, the total external force on the object is 2n and the direction is upward (opposite to the direction of the equal magnitude of f1 removed) [comprehensive formula: f = gsin37° -f2 = 12-10 = 2n].

First of all, the gravitational force is divided into the force 12n downward on the inclined plane and the force 16n downwards on the perpendicular inclined plane, because the object is stationary, the resultant force is zero, you can list such an equation: 2n 12n 10n frictional force, so the friction force is 4n, when it is zero, the same as above, the force analysis result is 12n-10n=2n downward direction, I used my mobile phone to call, it took up my own study time, I sure it is correct, I must do it.

g=4x10=40n

fxsin37= n=g-24n=16n f=un=fxcos37= f=32-f=24n a=f m=24 4=6m s square.

v = 2as under root = 2x6x4 = 4 under root 3