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Anonymous users2024-02-113v=(vb+vc)/2
Subtracting the two formulas yields: vc-va=4v
at=vc-va so a=4v t
a=s/t^2=(s7-s5)/(2t')^2=4/(2*1)^2=1m/s^2
s5=vt6+1/2at6^2-(vt5＋1/2at5^2)=v(t6-t5)+1/2a(t6^2-t5^2)
7.(2v) 2-v 2=2gs so the velocity v when landing'=2v=20m/s
t=2v'/g=2*20/10=4s
8.There is also a speed of 5m s when the screw cap is disengaged.
So the relative velocity at the beginning is 0
1 2gt 2=h so t = root number (
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Anonymous users2024-02-10a=2v/t
6.What kind of topic, I hate sports topics the most, (although the Olympiad has won awards).
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Anonymous users2024-02-091.Let the velocity at point A be V1, the acceleration time from Aa to B, and the time taken by T1 is only as follows.
v1+v1+t1*a)/2=v (1) (v1+t1*a+v1+t1*a+(t-t1)*a)/2=3v (2)
2v1+t1a+ta=3v (3) from 2
3)-(1) T*A=2V t=2V A
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Anonymous users2024-02-081. Suppose that the acceleration to the final speed is early v
v^2=2a1s1=2a2s2
t=v hydroponic a1+v a2
1/2at^2=s1+s2
Synopid: 1 a=1 a1+1 a2
2 Assume that the velocity is v
mg-180=ma1,a1=4m/s^2,360-mg=ma2,a2=2m/s^2
1)v=a1t1,t1=1s,v=4m/s
2)v^2=2a1s1=2a2s2
s1+s2=l,l=6m
3 pairs of wholes, land buried refers to f- (m1+m2+m3+...+mn)g=(m1+m2+m3+…+mn)a
For 1-k objects tk- (m1 + m2 + m3 + ....+mk）g=(m1+m2+m3+…+mk）a
Get tk=(m1+m2+m3+...+mk）f/(m1+m2+m3+…+mn)
4. Gravitational potential energy is converted into kinetic energy.
mgl+mg*2l=1/2mva^2+1/2mvb^2
vb=w*2l=2*wl=2va
va=√（6gl/5）,vb=2√（6gl/5）
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Anonymous users2024-02-071: Solution: Make a V-T diagram (make your own ha!!) ）
If car A travels T1 to reach the speed vt, and the total number of vehicles travels T, then car B travels T with A, and both cars travel S. (It's a bit cumbersome, hehe!) )
s=1 2*at square, s=1 2*t*vt
vt=at———1
vt=t1*a1,vt=(t-t1)*a2
t1=(ta2)/(a1+a2)
vt=t1*a1=(t*a1a2)/(a1+a2)=t*（1/a1+1/a2）——2
Proven by the formula! Phew
2: Solution: Sorry, the first question was not made, another day. %> This is the second question:
Force analysis diagram omitted).
a1=f m=(mg-fn) m=(300-180) 30=4m s square.
a1'=10-4=6m s square (direction down).
In the same way, we get: a2=f m=(fn-mg) m=(360-300) 30=2m s square.
a2'=10+2=12m s square (direction down).
l=1/2*a1'*1s square +1 2*a2'*2S square.
26m3: Solution: (using holistic analysis and isolation analysis).
If you find the tension of the rope between mx and m(x+1), you will see the first x blocks as a whole, and the blocks from (x+1) to nth as a whole.
Give a hint and do the math yourself
4: Solution: (figure omitted).
Set the angular velocity as
m square * 2l = mg, m = 3m
sqrt(g/2l)
va=ω*l
vb=ω*2l
o Ahhhh At 23:12, I'm going to sleep, and you can count it yourself, but I seem to have made a mistake, so add me as a friend, and I'll tell you about it later
As for these answers, you can make do with them!
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