13.( 1) Uniformly accelerated motion (2) Square of 4m/s (unit of acceleration) (3)1.6m/s.

14. a = V2-V1/t2=6-2m/s square (acceleration unit) x=v zero t+ 1/2a times the square of t to get t 1= 14 (irrelevant, omitted). So t

15.x 1=v zero t+ 1/2a times the square of t =7.5m x2=v zero t+ 1/2a times the square of t = 22.5m

V = x/t = 7.5/s

16.h = square of v/2g = 16/20m t 1=v-v zero/a = 0.4s.

H = h+64 = 16/20+642gh = squared at the end of v = 36m/s.

17.(2)x = 5/4 * 400+25(t-20)= 1000+ 15t t = 125s

(1) When t= 12, the maximum distance = 1090.

Physics ②1.b2.c3.b4.a5.a6.a7.a8.ab9.bc10.bc.

1 1.( 1) also write down the directions of the two chords, so that the node can be pulled to the O point.

(2) Root number 65

12.( 1)F=kx=6N

(2)F=kx=4N

(3)F=uFn=kx=4N

u=0.2

13.v=72000/3600=20m/s

A =- 1m/s square (acceleration unit)

F=m*a=- 1000N The direction of car pickup is the positive direction.

F=F +f= 1000N

F = m * aa =1000/1000 =1m/s squared (acceleration unit)

14. (1) v =10 times the root sign 10(2)t= 100.

15. (1) f1= 5/3mgf2 = 3/4mg (2) f = F2 to the left (3)m 1= 1.6kg, so it cannot exceed/kloc-.

Physics ③1.db2.bc3.CD4.acd5.ad6.ad7.CD8.c9.b10.b12.d.

13.5 14.( 1)F.C(2)A(3) ...

15.( 1)v = x/t = 264/ 12 = 22m/s(2)v = 10m/s

16.a=25 root number 3m/s x= 125 root number 3/2m.

17.f = cos 30 f = 15 radical 3N.

Fn= 135N

1 8. (1) u = 4/3 (2) a =10m/s squared (unit of acceleration) (3)v=9m/s,1,