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2 cot(2k) dk
IF = e ∫(1-3) - cot(2k)
IF = e
IF = e
IF = sin(2k)
2 ( ) In|sin(2k)|
IF = e ∫(1-n)P(x)
IF = e -2^ ∫-cot(2k)
IF = e ∫
1 2 In|sin(2k)| dk dx dk MAT 052 DELA VEGA, ANGELA KEIZY A. BSCE2 - 02 SITUATION 1 1.1 & 1. dh dk
- P(k)h = Q(k)h P(k) = -cot(2k) Q(k) = csc(2k) n = 3 n
y 1-n^ = ( ∫(1-n) Q(x) IF dx)
IF
h 1-3^ = ( ∫(1-3) csc(2k) sin(2k) dk)
sin (2k)
h = ( ∫-2csc(2k) sin(2k) dk)
sin (2k)
h -2^ = 1 ( -2k + C )
sin (2k)
h -2^ = -2k + C
sin (2k)
SITUATION 2 Pt = Poe kt f(t) = 200e (0.002t) 100,000 = 200e(0.002t) = e (0.002t)
In(500) = In(e (0.002t)) In(500) = 0.002t t = In(500)
t = 310.730 minutes
SITUATION 4 = inflow rate − outflow rate 2L min dV dt dV dt dV dt
= inflow rate − outflow rate
8g min min V(t) = 2dt V(t) = 2t + C V(0) = 8L V(0) = 2 (0) + C = 8 V(t) = 2t + 8 t = 20 minutes V(20) = 2(20) + 8 V(20) = 48L Inflow of chemical = ( ) ( ) = outflow of chemical = ( ) ( ) 2g L
4L
min A(t) V(t)
4L
dA dt
A(t) V(t) dA dt = 8 -^ 2 A(t) (2t + 8) dA dt
2 A(t) t + 4 dy dx
2 A(t) t + 4 dA
( dt )
2 A(t) t + 4
dA dt +^ A = 8
t + 4 IF = e IF = e IF = t + 4 1
∫ t +4 dt
In(t+4) (t+4) + A = 8 (t+4) dA dt d dt
∫ [(t+4) A] =∫^ 8 (t+4)
(t+4) A = 4 (t+4) + C^2 (0+4) A(0) = 4 (0+4) + C
A(0) = 32
(32) (4) = 4 (16) + C
(128) = 64 + C
C= 64
4(t+4) + 64 t + 4 A(t) = 2 2 4(t+4) + 64 t + 4
A(20) =
Finding the amount of chemical after 20 mins A(20) = 20 min 4(20+4) + 64 20 + 4
A(20) =
A(20) =
A(20) =
A(20) = 98. 667 g A(20) V(20)
Q(20) =
98.667g 48L
Q(20) =
Q(20) = 2.
g L Finding the concentration of chemical after 20 mins 2 2 SITUATION 3 To = 40 °F Ts = 70 °F 60 °F to 61 °F in 2 minutes (from 10:07 to 10:09) T(t) = Ts + (To - Ts) e-kt 61 = 70 + (60 - 70) e -k(2) -9 = -10 e-2k In(0.9) = -2k K = - In(0.9) 2 K = 0. 61 = 70 + (60 - 70) e -0.053t 60 = 70 - 30 e -0.053t -10 = -30e -0.053t (^1) = e -0.053t 3 In( 1 ) = -0.053t 3 t = - t = 20.729 minutes
In ( ) 4.1 & 4.
