8-2 ELECTRICAL СОMPU ТER FE AND PRACTICE PROBLEMS 11. A tank with a cross-sectional area of 12 m2 and a height of 10 m is filled with oil at a rate of 2.2 m3/min The density of the oil is 847 kg/m3. The oil leaks out of the tank from an open tap at the bottom of the tank The leak rate is 0.11t m3/min, where t is the time in elapsed minutes. Most nearly, what is the level of oil in the tank after the tank has been filled for 6 hr? 7. What is the solution to the following differential equation if x= - 1 at t- 0 , and dx/ dt= 0 at t 0? dax 1 8x 5 4- dt d2 4t e (A) -4t +4te (А) 5.2 m 5 -2cos 2t sin 2t) + 3 (B) (В) 6.6 m (C) 7.1 m -4t e 4t C) 4te 8 (D) 9.4 m 4t te -4t е 8 (D) x 8. In the following differential equation with the initial condition (0) = 12, what is the value of (2)? dx 4x 0 dt (A) 3.4 x 10-3 (В) 4.0 x 10-3 (С) 5.1 х 10-3 (D) 6.2 x 10-3 9. What are the three general Fourier coefficients for the sawtooth wave shown? f(t) 2 1 t -1 (А) аg — 0, а,р, 3D 0, b, Tn 1 -1 (B) ao аn — 0, bл 2 тп 1 (C) ao 1, an 1, b тп 1 1 (D) ao Иn 2 п 2 TTn 10. The values of an unknown function follow a Fibo- nacci number sequence. It is known that f(1) = 4 and f(2) 1.3. What is f(4)? (А) - 4.1 (В) 0.33 (C) 2.7 (D) 6.6 P PI ppi 2p ass.com 8-5 DIFFERENTIAL E QUATIONS Use the second-order difference equation 8. This is a first-order, linear, homogeneous differential equation with characteristic equation r+ 4 0. f(k) f(k-1) f(k 2) f(3) f(2)f(1) 1.34 4x 0 4t = 5.3 f(4) f(3)(2) = 5.31.3 x(0)= ne-4)(0) = 6.6 12 12 The answer is (D). 12e 4t x(2) 12e-4)(2) 11. The general equation for the unsteady-state mass balance is 12e-8 = 4.03 x 10 (4.0 x 103) min mout 'accumulation The answer is (B). The mass flow rates can be converted to volumetric flow rates 9. By inspection, f(t) t, with the period T 1. The angular frequency is Tin T out 'accumulation pQaccumulation in 2т 2т Qin Qaccumulation Qout 2T 1 Т The volume of oil accumulating in the tank changes with time The is average dV T Qinout (1/T)(t) dt (1/T)t 0 2 ао— dt 3 3 2.2 min 0.11t min 2 Since the cross-sectional area is constant, the volume of oil accumulating can be expressed in terms of the leak rate The general a term is (2/T)f(t)cos(nt) dt ат 3 m dh A. - 0.011t = 2.2 t cos (27nt) dt 2 dt min min 3 2.2 min 0.011t 0 = dh min A A dt The general b term is 3 m 0.011t 2.2 min Т b(2/T)f(t)sin(nwot)di min 12 m2 0.1833 m/min - 0.0009167t m/min 12 m2 t sin(2Tnt) dt = 2 1 The answer is (B). 10. The value of a number in a Fibonacci sequence is the sum of the previous two numbers in the sequence РPI Oрpі 2ра 8-6 FE ELECTRICAL AND COMPUTER PRACT1CE PROBLEMS Integrate both sides with respect to time. dh m |dt min 0.1833 min dt 0.0009167t dt 16h 0.0009167 t2 min m. 0.1833 h 2 min min = (0.1833 m)(6 h)(60- 0.0009167 min 60 h 2 = 6.585 m (6.6 m) The answer is (B). ppi2p as PPI Mathematics Mathematics Mathematics

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ISBN:9780190698614
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Publisher:Sadiku, Matthew N. O.
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attached question #11 and its solution but still I didn't understand it. can you explain it to me easier?

8-2
ELECTRICAL
СОMPU ТER
FE
AND
PRACTICE
PROBLEMS
11. A tank with a cross-sectional area of 12 m2 and a
height of 10 m is filled with oil at a rate of 2.2 m3/min
The density of the oil is 847 kg/m3. The oil leaks out of
the tank from an open tap at the bottom of the tank
The leak rate is 0.11t m3/min, where t is the time in
elapsed minutes. Most nearly, what is the level of oil in
the tank after the tank has been filled for 6 hr?
7. What is the solution to the following differential
equation if x=
- 1 at t- 0 , and dx/ dt= 0 at t 0?
dax
1
8x 5
4-
dt
d2
4t
e
(A)
-4t
+4te
(А) 5.2 m
5
-2cos 2t sin 2t) +
3
(B)
(В) 6.6 m
(C) 7.1 m
-4t
e
4t
C)
4te
8
(D) 9.4 m
4t
te
-4t
е
8
(D) x
8. In the following differential equation with the initial
condition (0) = 12, what is the value of (2)?
dx
4x 0
dt
(A) 3.4 x 10-3
(В) 4.0 x 10-3
(С) 5.1 х 10-3
(D) 6.2 x 10-3
9. What are the three general Fourier coefficients for
the sawtooth wave shown?
f(t)
2
1
t
-1
(А) аg — 0, а,р, 3D 0, b,
Tn
1
-1
(B) ao
аn — 0, bл
2
тп
1
(C) ao 1, an
1, b
тп
1
1
(D) ao
Иn
2
п
2
TTn
10. The values of an unknown function follow a Fibo-
nacci number sequence. It is known that f(1) = 4
and f(2) 1.3. What is f(4)?
(А) - 4.1
(В) 0.33
(C) 2.7
(D) 6.6
P PI ppi 2p ass.com
8-5
DIFFERENTIAL E QUATIONS
Use the second-order difference equation
8. This is a first-order, linear, homogeneous differential
equation with characteristic equation r+ 4 0.
f(k) f(k-1) f(k 2)
f(3) f(2)f(1) 1.34
4x
0
4t
= 5.3
f(4) f(3)(2) = 5.31.3
x(0)= ne-4)(0)
= 6.6
12
12
The answer is (D).
12e 4t
x(2) 12e-4)(2)
11. The general equation for the unsteady-state mass
balance is
12e-8
= 4.03 x 10 (4.0 x 103)
min mout
'accumulation
The answer is (B).
The mass flow rates can be converted to volumetric flow
rates
9. By inspection, f(t) t, with the period T 1. The
angular frequency is
Tin
T out
'accumulation
pQaccumulation
in
2т
2т
Qin
Qaccumulation
Qout
2T
1
Т
The volume of oil accumulating in the tank changes
with time
The
is
average
dV
T
Qinout
(1/T)(t) dt (1/T)t
0
2
ао—
dt
3
3
2.2
min
0.11t
min
2
Since the cross-sectional area is constant, the volume of
oil accumulating can be expressed in terms of the leak
rate
The general a term is
(2/T)f(t)cos(nt) dt
ат
3
m
dh
A.
- 0.011t
= 2.2
t cos (27nt) dt
2
dt
min
min
3
2.2
min
0.011t
0 =
dh
min
A
A
dt
The general b term is
3
m
0.011t
2.2
min
Т
b(2/T)f(t)sin(nwot)di
min
12 m2
0.1833 m/min - 0.0009167t m/min
12 m2
t sin(2Tnt) dt
= 2
1
The answer is (B).
10. The value of a number in a Fibonacci sequence is
the sum of the previous two numbers in the sequence
РPI Oрpі 2ра
8-6 FE
ELECTRICAL AND COMPUTER
PRACT1CE
PROBLEMS
Integrate both sides with respect to time.
dh
m
|dt
min
0.1833
min
dt
0.0009167t
dt
16h
0.0009167
t2
min
m.
0.1833
h
2
min
min
= (0.1833 m)(6 h)(60-
0.0009167
min
60
h
2
= 6.585 m
(6.6 m)
The answer is (B).
ppi2p as
PPI
Mathematics
Mathematics
Mathematics
Transcribed Image Text:8-2 ELECTRICAL СОMPU ТER FE AND PRACTICE PROBLEMS 11. A tank with a cross-sectional area of 12 m2 and a height of 10 m is filled with oil at a rate of 2.2 m3/min The density of the oil is 847 kg/m3. The oil leaks out of the tank from an open tap at the bottom of the tank The leak rate is 0.11t m3/min, where t is the time in elapsed minutes. Most nearly, what is the level of oil in the tank after the tank has been filled for 6 hr? 7. What is the solution to the following differential equation if x= - 1 at t- 0 , and dx/ dt= 0 at t 0? dax 1 8x 5 4- dt d2 4t e (A) -4t +4te (А) 5.2 m 5 -2cos 2t sin 2t) + 3 (B) (В) 6.6 m (C) 7.1 m -4t e 4t C) 4te 8 (D) 9.4 m 4t te -4t е 8 (D) x 8. In the following differential equation with the initial condition (0) = 12, what is the value of (2)? dx 4x 0 dt (A) 3.4 x 10-3 (В) 4.0 x 10-3 (С) 5.1 х 10-3 (D) 6.2 x 10-3 9. What are the three general Fourier coefficients for the sawtooth wave shown? f(t) 2 1 t -1 (А) аg — 0, а,р, 3D 0, b, Tn 1 -1 (B) ao аn — 0, bл 2 тп 1 (C) ao 1, an 1, b тп 1 1 (D) ao Иn 2 п 2 TTn 10. The values of an unknown function follow a Fibo- nacci number sequence. It is known that f(1) = 4 and f(2) 1.3. What is f(4)? (А) - 4.1 (В) 0.33 (C) 2.7 (D) 6.6 P PI ppi 2p ass.com 8-5 DIFFERENTIAL E QUATIONS Use the second-order difference equation 8. This is a first-order, linear, homogeneous differential equation with characteristic equation r+ 4 0. f(k) f(k-1) f(k 2) f(3) f(2)f(1) 1.34 4x 0 4t = 5.3 f(4) f(3)(2) = 5.31.3 x(0)= ne-4)(0) = 6.6 12 12 The answer is (D). 12e 4t x(2) 12e-4)(2) 11. The general equation for the unsteady-state mass balance is 12e-8 = 4.03 x 10 (4.0 x 103) min mout 'accumulation The answer is (B). The mass flow rates can be converted to volumetric flow rates 9. By inspection, f(t) t, with the period T 1. The angular frequency is Tin T out 'accumulation pQaccumulation in 2т 2т Qin Qaccumulation Qout 2T 1 Т The volume of oil accumulating in the tank changes with time The is average dV T Qinout (1/T)(t) dt (1/T)t 0 2 ао— dt 3 3 2.2 min 0.11t min 2 Since the cross-sectional area is constant, the volume of oil accumulating can be expressed in terms of the leak rate The general a term is (2/T)f(t)cos(nt) dt ат 3 m dh A. - 0.011t = 2.2 t cos (27nt) dt 2 dt min min 3 2.2 min 0.011t 0 = dh min A A dt The general b term is 3 m 0.011t 2.2 min Т b(2/T)f(t)sin(nwot)di min 12 m2 0.1833 m/min - 0.0009167t m/min 12 m2 t sin(2Tnt) dt = 2 1 The answer is (B). 10. The value of a number in a Fibonacci sequence is the sum of the previous two numbers in the sequence РPI Oрpі 2ра 8-6 FE ELECTRICAL AND COMPUTER PRACT1CE PROBLEMS Integrate both sides with respect to time. dh m |dt min 0.1833 min dt 0.0009167t dt 16h 0.0009167 t2 min m. 0.1833 h 2 min min = (0.1833 m)(6 h)(60- 0.0009167 min 60 h 2 = 6.585 m (6.6 m) The answer is (B). ppi2p as PPI Mathematics Mathematics Mathematics
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