ED 2 Semestre UNIP Engenharia

RESOLUÇÕES

01-

100 KG- agricultura atual

20 kg- N,10 kg- Oxi. P, 10 kg-Oxi. K

300 kg- terra nossa

30 kg-N, 30kg- Oxi. P, 60 kg- Oxi. K

400 kg- (misturar os 100 kg-agri. Atual + 300 Kg- terra nossa)

50 kg- N, 40 kg- Oxi. P, 70 kg- Oxi. K

50/400*100=12,5%

40/400*100=10%

70/400*100=17,5% RESP.: B

02-

(35,0);(5,60)

T=aL+b

35=a*0+b; b=35

5=a*60+35; 60a=-35+5; a=-30/60; a=-0,5

T=35-0,5L RESP.: A

03-

P/ o objeto atingir o chão, h=0

H=-4,9t²+49; -4,9t²+49=0; -4,9t²=-49; t²=10; t=√10; t=3,16; t≈3,2s. RESP.:D

04-

IB=t²-24t+143

IB’=2t-24

Minimo

IB’=0; 2t-24=0; 2t=24; t=12hIB(12)= 12²-24.12+143

IB(12)=-1 RESP.:E

05-

A=(-2,3) ; B(1,-4)

AB=(1-(-2),-4-3); AB=(3,-7) RESP.:E

06-

u.v=|u|.|v|.cosⱷ; u.v=6*9*cos150°; u.v=54*(-√3/2); u.v=-27√3; O cosseno de 150° é o mesmo cosseno de 30° com sentido oposto, por estar no segundo quadrante. Cos30°=√3/2;

cos150°=-√3/2 RESP.:B

07- B

08- C

09-

V(t)=15t²-750t+9000

V(3)=15*3²-750*3+9000; v(3)=6885L RESP.: A

10-

V(t)=15t²-750t+9000

V’(t)=15*2t-750; v’(3)=30*3-750; v’(3)=-660L/h RESP.:E

11-

V(t)=-4,5t²+18t

V’(t)=2(-4,5)t+18; v’(t)= -9t+18; v’(t)=0; -9t+18=0; 9t=18; t=2s

V(2)=-4,5(2)²+18*2; v(2)= 18 m/s RESP.:D

12-

I. |U|=√((-3)²+4²+0²); |U|=√25; |U|=5 V

II. Versor => u/|u|=(-3/5,4/5,0); u/|u|= (-0,6;0,8;0) V

III. u=15(-0,6;0,8;0); u=(-9,12,0)

-u=15(-0,6;0,8;0); -u=(9,-12,0) V

RESP.: A

13-

W=αu+βv

(-17,12) = α(-2,0)+β(3,-4); (-17,12) = (-2α,0) + (3β,-4β); (-17,12)=(-2α+3β,-4β)

-17=-2α+3β; 12=-4β; β=-3; -17=-2α+3(-3); -2α=-8; α=4 RESP.:A

14-

AO+DE+FG+PL-IH PL=IH

AO+OD+DP=AP RESP.: A

15-

AQ=AC+CG+(2/3)*GH; AQ=AC+AE+(2/3)*AB RESP.:E

16-

I. 2*(1,-2) -4*(-4,0)= (2+16, -4); (18, -4) V

II. (1, -2) + (-4, 0) = (-3, -2); √((-3)²+(-2)²); √(9+4); √13 V

III. 1/(-4)=-2/0 F

RESP.:D

17-

(X+12)/6=3/9; 9(X+12)=3*6; 9X+108=18; 9X=-90; X=-10 RESP.:B

18-

S=(3,-6); |S|=√(3²+(-6)²); |S|=√45; |S|=√(5*3²); |S|=3√5 V RESP.:D

19-

A=(-1, 3) ; B=(0,