UiTM – Faculty of Applied Science
Test 3 – Set B - PHY430 – AS2232/CS2273 – 21.6.12
SOLUTIONS
Answer all the questions (g = 9.80 m/s2)
1.
a)
State and explain Continuity Equation.
Continuity Equation: A1v1 = A2v2 = constant
The product of the area and the fluid speed at all points along a pipe is constant
for an incompressible fluid.
Volume : symbol capital V
speed: symbol small v
𝑑𝑉
The product Av is called the volume flow rate ( ) which has the dimension of
𝑑𝑡
volume per unit time.
b)
Water flowing through a garden hose of diameter 0.030 m fills a 0.025-m3 bucket
in 90 s. Determine
i) the rate at which the water fills the bucket
𝑑𝑉
𝑑𝑡
=
0.025
90
𝑚3
= 2.8 × 10−4
𝑠
ii) the speed of the water leaving the end of the hose
𝐴1 𝑣1 =
𝑣1 =
𝑑𝑉
𝑑𝑡
2.8 × 10−4
0.030 2
𝜋(
)
2
= 0.40
𝑚
𝑠
iii) the speed of the water leaving the end of the hose if the cross-sectional area
of the end of the hose is reduced to one-third its original cross-sectional area.
𝐴2 𝑣2 =
𝑣2 =
𝑑𝑉
𝑑𝑡
2.8 × 10−4
1
0.030 2
( )𝜋(
)
3
2
= 1.2
𝑚
𝑠
2.
A sinusoidal travelling wave on a string is represented by y = 0.15 sin (0.80x + 50t)
where y and x are in meters and t is in seconds. For this wave determine
Compare
y = 0.15 sin (0.80x + 50t)
y = A sin (kx + t)
k = 0.80 m1 and  = 50 rad/s
a) the wavelength
𝑘=
2𝜋
2𝜋
2𝜋
⟹𝜆=
=
= 7.9 𝑚
𝜆
𝑘
0.80
b) frequency
𝜔 = 2𝜋𝑓 ⟹ 𝑓 =
 50
=
= 8.0 𝑠 −1
2 2
c) velocity of the wave (magnitude and direction)
𝑣 = 𝜆𝑓 = (7.9)(8.0) = 63
𝑚
𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡
𝑠
d) maximum and minimum speed of the particles of the string.
𝑣𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 =
𝑑𝑦
= 0.15 𝑐𝑜𝑠(0.80𝑥 + 50𝑡)(50)
𝑑𝑡
= 7.5 𝑐𝑜𝑠(0.80𝑥 + 50𝑡)
𝑣𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = 0 𝑤ℎ𝑒𝑛 𝑐𝑜𝑠(0.80𝑥 + 50𝑡) = 0
𝑚
𝑣𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 = 7.5
𝑤ℎ𝑒𝑛 𝑐𝑜𝑠(0.80𝑥 + 50𝑡) = 1
𝑠
3.
a)
State the meaning of
i)
specific heat capacity
Heat needed to increase the temperature of a 1 kg substance by 1 K.
𝐽
The unit is 𝑘𝑔∙𝐾.
ii)
latent heat of vaporization.
Heat needed to change a 1 kg substance from liquid to gas without any
temperature change.
The unit is
𝐽
.
𝑘𝑔
b) Calculate the amount of thermal energy needed to change a 0.720-kg ice at 10C
to water at 15C.
Q needed =
Q to increase the temperature of ice from 10C to 0C
+ Q to melt all the ice at 0C
+ Q to increase the temperature of water from 0C to 15C
=
mi ci i + mi Lf + mw cw w
=
(0.720)(2100)(0  10) + (0.720)( 3.33  105) +
(0.720)(4200)(15  0)
=
3.00  105 J
Specific heat of ice = 2100 J/kg.K
Specific heat of water = 4200 J/kg.K
Latent heat of fusion of water = 3.33  105 J/kg