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Vinvika [58]
3 years ago
8

1) A cylinder has a volume of 20 cubic feet. What is that volume in cubic inches? (1 ft = 12 in) ​

Engineering
1 answer:
hram777 [196]3 years ago
7 0
Before calculating the volume of this cylinder, we must either convert the diameter to feet or the height to inches. I'll do the latter, to avoid getting a very small number.
Recall that there are
12
inches in
1
foot. Thus, we can state:
12
in.
1
ft.
=
x
4
ft.
x
=
12
in.
×
4
ft.
1
(ft.)
x
=
48
in.
Hence, the cylinder has a height of
48
inches.
Now, we can apply the formula for volume of a cylinder to effectuate our calculation. The formula in question is
V
=
a
base
×
h
, or
V
=
r
2
π
×
h
.
However, we know our diameter but we don't know our radius. As you probably know, the diameter is linked to the radius b the formula
d
=
2
r
. Solving for
r
and substituting:
r
=
d
2
r
=
12
2
r
=
6
∴
The radius of the cylinder measures
6
inches.
V
=
r
2
π
×
h
V
=
6
2
π
×
48
V
=
(
36
×
48
)
π
V
=
1728
π
in
2
Note that this answer is in exact value. Rounded to two decimal places, the volume is
5428.67
in
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*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t
riadik2000 [5.3K]

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

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δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

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Thus, ε_x = δ/L = 0.03/4 = 0.0075

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3 years ago
Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,
Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take p_o = 880 kg/m³ and p_{glycerol = 1260 kg/m³    

 

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

Given that;

Inlet velocity of Glycerin, V_A = 6 m/s

Inlet velocity of oil, V_B = 3 m/s  

Density velocity of glycerin, p_{glycerol = 1260 kg/m³

Density velocity of glycerin, Take p_o = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, d_A = 100 mm = 0.1 m

Diameter of oil pipe, d_B = 80 mm = 0.08 m

Diameter of outlet pipe d_C = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

Q_A + Q_B = Q_C

A_Av_A + A_Bv_B = A_Cv_C

π/4 × (d_A)²v_A + π/4 × (d_B )²v_B = π/4 × (d_C)²v_C

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²v_C

0.04712 + 0.0150796 = 0.0113097v_C

0.0621996 = 0.0113097v_C

v_C = 0.0621996 / 0.0113097

v_C  = 5.5 m/s

Now we apply the mass flow rate condition

m_A + m_B = m_C

p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C  

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p  

72.6412 = 0.06220335p    

p = 72.6412 / 0.06220335

p =  1167.8 kg/m³  

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

4 0
3 years ago
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