a) Terminal Velocity in Water
Given Data:
Diameter of grain (
𝑑
d):
4.318
×
1
0
−
5
m
4.318×10
−5
m
Density of grain (
𝜌
𝑠
ρ
s
):
1443.3
kg/m
3
1443.3 kg/m
3
Density of water (
𝜌
𝑤
ρ
w
):
1000
kg/m
3
1000 kg/m
3
Viscosity of water (
𝜇
𝑤
μ
w
):
1.1
×
1
0
−
3
N
\cdotp
s/m
2
1.1×10
−3
N\cdotps/m
2
Conversions:
Convert diameter to radius:
𝑟
=
𝑑
2
=
4.318
×
1
0
−
5
2
=
2.159
×
1
0
−
5
m
r=
2
d
=
2
4.318×10
−5
=2.159×10
−5
m
Calculate Terminal Velocity in Water:
Substitute the values into the formula:
𝑣
𝑡
=
2
⋅
(
2.159
×
1
0
−
5
)
2
⋅
(
1443.3
−
1000
)
⋅
9.81
9
⋅
1.1
×
1
0
−
3
v
t
=
9⋅1.1×10
−3
2⋅(2.159×10
−5
)
2
⋅(1443.3−1000)⋅9.81
Calculate:
𝑣
𝑡
=
2
⋅
4.66
×
1
0
−
10
⋅
443.3
⋅
9.81
9.9
×
1
0
−
3
v
t
=
9.9×10
−3
2⋅4.66×10
−10
⋅443.3⋅9.81
𝑣
𝑡
=
4.51
×
1
0
−
7
9.9
×
1
0
−
3
≈
4.55
×
1
0
−
5
m/s
v
t
=
9.9×10
−3
4.51×10
−7
≈4.55×10
−5
m/s
b) Terminal Velocity in Air
Given Data:
Density of air (
𝜌
𝑎
ρ
a
):
1.31
kg/m
3
1.31 kg/m
3
Viscosity of air (
𝜇
𝑎
μ
a
):
9.0
×
1
0
−
6
N
\cdotp
s/m
2
9.0×10
−6
N\cdotps/m
2
Calculate Terminal Velocity in Air:
Substitute the values into the same formula:
𝑣
𝑡
=
2
⋅
(
2.159
×
1
0
−
5
)
2
⋅
(
1443.3
−
1.31
)
⋅
9.81
9
⋅
9.0
×
1
0
−
6
v
t
=
9⋅9.0×10
−6
2⋅(2.159×10
−5
)
2
⋅(1443.3−1.31)⋅9.81
Calculate:
𝑣
𝑡
=
2
⋅
4.66
×
1
0
−
10
⋅
1442
⋅
9.81
8.1
×
1
0
−
5
v
t
=
8.1×10
−5
2⋅4.66×10
−10
⋅1442⋅9.81
𝑣
𝑡
=
1.38
×
1
0
−
6
8.1
×
1
0
−
5
≈
0.017
m/s
v
t
=
8.1×10
−5
1.38×10
−6
≈0.017 m/s
Summary:
Terminal Velocity in Water: Approximately
4.55
×
1
0
−
5
4.55×10
−5
m/s
Terminal Velocity in Air: Approximately
0.017
0.017 m/
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