Do not actually solve the problem numerically or algebraically, just pick the one equation and define the relevant knowns and single unknown. Don’t forget to include direction when called for by a vector variable 12) The air conditioner removes 2.7 kJ of heat from inside a house with 450 m3 of air in it. At a typical air density of 1.3 kg/m3 that means 585 kg of air. If the specific heat of air is 1.01 kJ/(kg oC), by how much would this cool the house if no heat got in through the rest of the house during that time?
Do not actually solve the problem numerically or algebraically, just pick the one equation and define the relevant knowns and single unknown. Don’t forget to include direction when called for by a vector variable 12) The air conditioner removes 2.7 kJ of heat from inside a house with 450 m3 of air in it. At a typical air density of 1.3 kg/m3 that means 585 kg of air. If the specific heat of air is 1.01 kJ/(kg oC), by how much would this cool the house if no heat got in through the rest of the house during that time?
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Do not actually solve the problem numerically or algebraically, just pick the one equation and define the relevant knowns and single unknown. Don’t forget to include direction when called for by a vector variable
12) The air conditioner removes 2.7 kJ of heat from inside a house with 450 m3 of air in it. At a typical air density of 1.3 kg/m3 that means 585 kg of air. If the specific heat of air is 1.01 kJ/(kg oC), by how much would this cool the house if no heat got in through the rest of the house during that time?

Transcribed Image Text:x(t) = x0 + Voxt + ½ axt²
y(t) = yo + Voyt + ½ ayt²
Vx (t) = Vox? + 2ax(x-xo)
Vy (t) = Voy + 2a,(y-yo)
v(t) = dx/dt
a(t) = d?x/dt?
Vx(t) = Vox + axt
Vy(t) = Voy + ayt
<V> = (vo + V1) / 2
<a> = (V1 – Vo)/(tj – to)
r = x² + y?
(Vo + V1) / 2 = Ar/At
a(t) = dv/dt
x = rcose
<V> = Ar/At
<a> = Av/At
|A| = (A,? + A,?)2
 = A
y = rsin0
a = ac(-f) + arê®
ac = Vr/r
a. = m²r
VT = (2nr)/T
ac = 4n°r/T?
ar = dvf/dt
VT = or
f = o/2n
f = 1/T
rP/A = rP/B + rB/A
VP/A = VP/B + VB/A
ap/A = ap/B + aB/A
m, m2
EF¡ = ma
F12 = -F21
= G
(-12)
fs,max = lsFN
F = mv²/r
fk = µkFN
Fe = mac
Fdrag
= -by
Fdrag = ½ DpAv² (-8)
2mg
Vt = mg/b
Vt =
dW; = F; · dr
Wi = F; · Ar
Ug = mgy
AUint = fgAs
n = Pout/Pin (X 100%)
V DpA
K = ½ mv?
Usp = ½ k(x-xrel)?
P = dW/dt
AK = ½ m (v1²-vo²)
Fsp = -k (x-Xrel)
P = dU/dt
= dp/dt
ΣW ΔΚ
P = F•v
<ΣF-Δp/Δt
p= mv
miv10 + m2v20 = m¡v11 + m2V21
½ miV10? + ½ m2V20? = ½ m¡V11? + ½ m2v21?
m¡V10 + m2v20 = (m¡+m2)VTI
(mi+m2)VT0 = m¡Vj1 + m2V21
(m2-m1
m,+m2
(Em;) Zem = E(m;z;)
(mi-m2) V10 +
m1+m2/
2m2
2m1
V11
V20
V21 =
V20 +
|V10
\m,+m2/
(Σm) xem- Σ(mx )
Av, = +Ve In(M/M¡)
0 = 0o + @ot + ½ at²
\m,+m2/
(Σm ) yem Σ(my )
Fth = v(dm/dt)
mtotalXCM =
o? = 00? + 2a(0–0)
@ = Wo + at
<a> = A@/At
<@> = AÐ/At
Vcm = ro
S = r0
a(t) = d²0/dt?
T = rleverF
TAB = -TBA
I= Iem + Md²
@(t) = d0/dt
a(t) = d@/dt
T = rFsino
Et = dL/dt
I = BMR?
L =r xp
T =r x F
dW; = t;· de
I= E m,r;?
Op = (mgrsin0)/(Io)
Στ-Ια
Krot = ½ Io?
rjever = rsino
I010 + I2020 = I011 + I2021
(I1+I2)@ro = I10i1 + I2@21
L = Io
ΔL TΔt
I010 + I2020 = (I1+I2)@T1
I10010 = I1101
1/2
![F
ΔF
ΔΡ
Y =
B =
B =
AV
S =
Vo
x(t) = Xmax cOs(@t + 0)
T= 27 (L/g)'2
@ = (k/m)'2
T= 27 (I/mgd)"2
Vmax = OXmax
amax = 0ʻxmax
y(x,t) = ymax Ssin(kx + @t + þ)
v = f.
a?y
Odamped = ( (k/m) – (b/2m)² )'/²
y(x,t) = ymax sin(kx - ot + ¢)
2 = 2n/k
v = @/k
1 д?у
v = (Fr/u)2
µ = m/L
<Pwave> = ½ uymax-@ʻv
əx²
v² at2
ΔΡ
%3D
max
Bksmax AP,
= pwvsSmax
Vsound =
Vsound = 343 m/s
max
TC
Vsound =
(331-) 1+
Vsound 331 m/s + (0.60 m/s°C) Tc
273 °C
APmax
I =
2ρν
10 dB = 1 B
<P>
B = log ()
P = pvosmax A sin°(kx-@t)
I, = 10-12 W/m² (exactly)
I =
%3D
4tr2
fobserver = fsource
(Vsound+Vobserver
sin(A) + sin(B) = 2 cos (-4):
sin ()
Vsound-Vsource
0 = 2tn (const)
0 = T(2n+1) (dest)
An = 4L/(2n-1)
fn = (2n-1)v/4L
AV = V.B(T1-To) AL=La(T1-To)
Q = mLv
AEint = W + Q
n = 0, ±1, ±2,..
An = 2L/n
fn = nv/2L
n = 1, 2, 3, ...
Af = |f1 – f2|
B= 3a
Q = mLf
AEcyele
PV = nRT
Q = mcAT
W = - SP dV
AT = Tinal - Tinitial
Wisobaric = -PAV
Vfinal
\Vinitial-
= 0
Wisovolumetric = 0
Qadiabatic = 0
Qisothermal
= nRT In
R = Ax/k
4
Pradiated = GAETK*
TK = Tc + 273.15
Ktot = ( ½ NKBT) * degrees of freedom (dof)
8kgT
P = kA|dT/dx|
Pnet = GAe (Tsource – Tobject")
½ mo<v?> = ½ kgT
3kgT
kB = R/NA
2kgT
VRMS =
Vavg
Vmost likely
mo
Timo
V mo
3RT
8RT
2RT
VRMS =
Vavg
Vmost likely
M
M
AEint = Q = nCyAT Q=nCpAT
P¡V = P2V2Y
CoP = |Q/|W|
dS = dQreversible/T ASfree expand = nR In(V2/V1)
Ср 3D Су + R
Y = Cp/Cy
n = |W]/lQH|
Cy = ½ R * dof
T¡V;! = T2V;*1
n = 1 – (IQc/IQH|)
Notto = 1– (V2/V)-!
= 0
NCarnot = 1 – (T/TH)
ASCarnot
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fb1edcb-e378-4007-8f8b-0be0a89cb7b5%2F0d68606f-cde1-47c0-b10c-50f30f2c6528%2Fzw2ig4r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:F
ΔF
ΔΡ
Y =
B =
B =
AV
S =
Vo
x(t) = Xmax cOs(@t + 0)
T= 27 (L/g)'2
@ = (k/m)'2
T= 27 (I/mgd)"2
Vmax = OXmax
amax = 0ʻxmax
y(x,t) = ymax Ssin(kx + @t + þ)
v = f.
a?y
Odamped = ( (k/m) – (b/2m)² )'/²
y(x,t) = ymax sin(kx - ot + ¢)
2 = 2n/k
v = @/k
1 д?у
v = (Fr/u)2
µ = m/L
<Pwave> = ½ uymax-@ʻv
əx²
v² at2
ΔΡ
%3D
max
Bksmax AP,
= pwvsSmax
Vsound =
Vsound = 343 m/s
max
TC
Vsound =
(331-) 1+
Vsound 331 m/s + (0.60 m/s°C) Tc
273 °C
APmax
I =
2ρν
10 dB = 1 B
<P>
B = log ()
P = pvosmax A sin°(kx-@t)
I, = 10-12 W/m² (exactly)
I =
%3D
4tr2
fobserver = fsource
(Vsound+Vobserver
sin(A) + sin(B) = 2 cos (-4):
sin ()
Vsound-Vsource
0 = 2tn (const)
0 = T(2n+1) (dest)
An = 4L/(2n-1)
fn = (2n-1)v/4L
AV = V.B(T1-To) AL=La(T1-To)
Q = mLv
AEint = W + Q
n = 0, ±1, ±2,..
An = 2L/n
fn = nv/2L
n = 1, 2, 3, ...
Af = |f1 – f2|
B= 3a
Q = mLf
AEcyele
PV = nRT
Q = mcAT
W = - SP dV
AT = Tinal - Tinitial
Wisobaric = -PAV
Vfinal
\Vinitial-
= 0
Wisovolumetric = 0
Qadiabatic = 0
Qisothermal
= nRT In
R = Ax/k
4
Pradiated = GAETK*
TK = Tc + 273.15
Ktot = ( ½ NKBT) * degrees of freedom (dof)
8kgT
P = kA|dT/dx|
Pnet = GAe (Tsource – Tobject")
½ mo<v?> = ½ kgT
3kgT
kB = R/NA
2kgT
VRMS =
Vavg
Vmost likely
mo
Timo
V mo
3RT
8RT
2RT
VRMS =
Vavg
Vmost likely
M
M
AEint = Q = nCyAT Q=nCpAT
P¡V = P2V2Y
CoP = |Q/|W|
dS = dQreversible/T ASfree expand = nR In(V2/V1)
Ср 3D Су + R
Y = Cp/Cy
n = |W]/lQH|
Cy = ½ R * dof
T¡V;! = T2V;*1
n = 1 – (IQc/IQH|)
Notto = 1– (V2/V)-!
= 0
NCarnot = 1 – (T/TH)
ASCarnot
%3D
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