Answer:
See explanation
Explanation:
See the document for the complete FBD and the introductory part of the solution.
Static Balance ( Sum of Forces = 0 ) in all three directions
∑[tex]F_G_X = W - G_x = 0[/tex]
[tex]G_X = W = 1500 lb[/tex]
∑[tex]F_G_Y = P - G_Y = 0[/tex]
[tex]G_Y = P = -10,800 lb[/tex]
∑[tex]F_G_Z = - G_Z = 0[/tex]
Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
[tex]M_G = r_O_G x F_O[/tex]
Where,
r_OG: The vector from point O to point G
F_OG: The force vector at point O
- The vector ( r_OG ) and ( F_OG ) can be written as follows:
[tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]
[tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
[tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
[tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]
[tex]V = P = -10,800 lb[/tex]
- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-
σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]
Where,
A: The area of pipe cross section
[tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]
z*: The distance of point "C" along z-direction from central axis ( x )
[tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]
I_YY: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
y*: The distance of point "C" along y-direction from central axis ( x )
[tex]y* = 0[/tex]
- The normal stress ( σ_x ) becomes:
σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]
Where,
c: The radial distance from central axis ( x ) and point "C".
[tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]
J: The polar moment of inertia of the annular cross section of pipe:
[tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]
Q: The first moment of area for point "C" = semi-circle
[tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]
I: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
t: The effective thickness of thin walled pipe:
[tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]
- The planar stress is ( τ_xy ) becomes:
τ_xy = [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]
- The principal stresses at point "C" can be determined from the following formula:-
σ_x = 15.55 ksi, σ_y = 0 ksi , τ_xy = 52.4 ksi
σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]
σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]
- The angle of maximum plane stress ( θ ):
θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]
Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.
A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.
Answer:
The correct answer to the following question will be "1.23 mm".
Explanation:
The given values are:
Average normal stress,
[tex]\sigma=200 \ MPa[/tex]
Elastic module,
[tex]E = 77 \ GPa[/tex]
Length,
[tex]L = 570 \ mm[/tex]
To find the deformation, firstly we have to find the equation:
⇒ [tex]\delta=\Sigma\frac{N_{i}L_{i}}{E \ A_{i}}[/tex]
⇒ [tex]=\frac{P(\frac{L}{H})}{E(bt)} +\frac{P(\frac{L}{2})}{E (bt)(\frac{2}{3})}+\frac{P(\frac{L}{H})}{Ebt}[/tex]
On taking "[tex]\frac{PL}{Ebt}[/tex]" as common, we get
⇒ [tex]=\frac{\frac{PL}{Ebt}}{[\frac{1}{4}+\frac{3}{4}+\frac{1}{4}]}[/tex]
⇒ [tex]=\frac{5PL}{HEbt}[/tex]
Now,
The stress at the middle will be:
⇒ [tex]\sigma=\frac{P}{A}[/tex]
⇒ [tex]=\frac{P}{(\frac{2}{3})bt}[/tex]
⇒ [tex]=\frac{3P}{2bt}[/tex]
⇒ [tex]\frac{P}{bt} =\frac{2 \sigma}{3}[/tex]
Hence,
⇒ [tex]\delta=\frac{5 \sigma \ L}{6E}[/tex]
On putting the estimated values, we get
⇒ [tex]=\frac{5\times 200\times 570}{6\times 77\times 10^3}[/tex]
⇒ [tex]=\frac{570000}{462000}[/tex]
⇒ [tex]=1.23 \ mm[/tex]
In a hydroelectric power plant, water enters the turbine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity (m/s) to which water can be accelerated by the nozzles before striking the turbine blades.
Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
Answer:
[tex]V2 = 37.417ms^{-1}[/tex]
Explanation:
Given the following data;
Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.
Nozzle outlets = 100kPa = 100000pa.
Density of water = 1000kg/m³.
We would apply, the Bernoulli equation between the inlet and outlet;
[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]
Where, V1 is approximately equal to zero(0).
Z[tex]z_{1} = z_{2}[/tex]
Therefore, to find the maximum velocity, V2;
[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]
[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]
[tex]V2 = \sqrt{2(800-100)}[/tex]
[tex]V2 = \sqrt{2(700)}[/tex]
[tex]V2 = \sqrt{1400}[/tex]
[tex]V2 = 37.417ms^{-1}[/tex]
Hence, the maximum velocity, V2 is 37.417m/s
list everything wrong with 2020
Answer:
George Floyd (BLACK LIFES MATTER)
C O V I D - 19
Quarantine
no sports
wearing a mask
and a whole lot of other stuff
Explanation:
A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 4200 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assume uniformly accelerated motion. determine the number of revolutions that the motor executes
(a) in reaching its rated speed,
(b) in coating to rest.
Answer:
a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]
Explanation:
a) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]
[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 175\,rev[/tex]
b) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]
[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 2450\,rev[/tex]
Purely resistive loads of 24 kW, 18 kW, and 12 kW are connected between the neutral
and the red, yellow and blue phases respectively of a 3-0, four-wire system. The line
voltage is 415 V. Calculate:
i. the current in each line conductor (i.e., IR ,Iy and IB); and
ii. the current in the neutral conductor.
Answer:
(i) IR = 100.167 A Iy = 75.125∠-120 IB = 50.083 ∠+120 (ii) IN =43.374∠ -30°
Explanation:
Solution
Given that:
Three loads 24 kW, 18 kW, and 12 kW are connected between the neutral.
Voltage = 415V
Now,
(1)The current in each line conductor
Thus,
The Voltage Vpn = vL√3
Gives us, 415/√3 = 239.6 V
Then,
IR = 24 K/ Vpn ∠0°
24K/239.6 ∠0°= 100.167 A
For Iy
Iy = 18k/239. 6
= 75.125A
Thus,
Iy = 75.125∠-120 this is as a result of the 3- 0 system
Now,
IB = 12K /239.6
= 50.083 A
Thus,
IB is =50.083 ∠+120
(ii) We find the current in the neutral conductor
which is,
IN =Iy +IB +IR
= 75.125∠-120 + 50.083∠+120 +100.167
This will give us the following summation below:
-37.563 - j65.06 - 25.0415 +j 43.373 + 100.167
Thus,
IN = 37.563- j 21.687
Therefore,
IN =43.374∠ -30°
Participating in extracurricular activities in high school helps:
Answer:
Develop social skills
Explanation:
Answer:
strengthen your college applications
Explanation:
Describe with an example how corroded structures can lead to environment pollution?
The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has rotated 4 revolutions starting from rest. The wheel has a radius of gyration of kO = 0.3 m.
Image of wheel is missing, so i attached it.
Answer:
ω = 14.95 rad/s
Explanation:
We are given;
Mass of wheel; m = 20kg
T = 20 N
k_o = 0.3 m
Since the wheel starts from rest, T1 = 0.
The mass moment of inertia of the wheel about point O is;
I_o = m(k_o)²
I_o = 20 * (0.3)²
I_o = 1.8 kg.m²
So, T2 = ½•I_o•ω²
T2 = ½ × 1.8 × ω²
T2 = 0.9ω²
Looking at the image of the wheel, it's clear that only T does the work.
Thus, distance is;
s_t = θr
Since 4 revolutions,
s_t = 4(2π) × 0.4
s_t = 3.2π
So, Energy expended = Force x Distance
Wt = T x s_t = 20 × 3.2π = 64π J
Using principle of work-energy, we have;
T1 + W = T2
Plugging in the relevant values, we have;
0 + 64π = 0.9ω²
0.9ω² = 64π
ω² = 64π/0.9
ω = √64π/0.9
ω = 14.95 rad/s
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa. After being compressed, the air is at 450 °C. Determine
(a) the final pressure in [MPa],
(b) the increase in total internal energy in [kJ], and
(c) the total work required in [kJ].
Note that for air R-287 J/kg.K and c.-716.5 J/kg.K, and ?-
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Given in the following v(t) signal.
a. Find the first 7 harmonics of the Fourier series function in cosine form.
b. Plot one side spectrum
c. Find the first 7 harmonics of the Fourier series function in exponential form.
d. Plot two side spectrum Given in the following v(t) signal.
Answer:
Check the v(t) signal referred to in the question and the solution to each part in the files attached
Explanation:
The detailed solutions of parts a to d are clearly expressed in the second file attached.
In contouring, it is necessary to measure position and not velocity for feedback.
a. True
b. False
In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
a. True
b. False
Job shop is another term for process layout.
a. True
b. False
Airplanes are normally produced using group technology or cellular layout.
a. True
b. False
In manufacturing, value-creating time is greater than takt time.
a. True
b. False
Answer:
(1). False, (2). True, (3). False, (4). False, (5). True.
Explanation:
The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.
(1). In contouring, it is necessary to measure position and not velocity for feedback.
ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.
(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
ANSWER: a=> True.
(3). Job shop is another term for process layout.
ANSWER: b => False
JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.
(4). Airplanes are normally produced using group technology or cellular layout.
ANSWER: b => False.
(5). In manufacturing, value-creating time is greater than takt time.
ANSWER: a => True.
A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of 62 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a design factor of 1.5, what dimension should the square cross section have
Answer:
The dimension of the square cross section is = 30mm * 30mm
Explanation:
Before proceeding with the calculations convert MPa to Kpsi
Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi
the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut
at 10^6 cycles" for 111.65 Kpsi = f ( fatigue strength factor) = 0.83
To calculate the endurance limit use Se' = 0.5 Sut since Sut < 1400 MPa
Se'( endurance limit ) = 0.5 * 770 = 385 Mpa
The surface condition modification factor
Ka = 57.7 ( Sut )^-0.718
Ka = 57.7 ( 770 ) ^-0.718
Ka = 0.488
Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one
The endurance limit at the critical location of a machine part can be expressed as :
Se = Ka*Kb*Se'
Se = 0.488 * 0.85 * 385 = 160 MPa
Next:
Calculating the constants to find the number of cycles
α = [tex]\frac{(fSut)^2}{Se}[/tex]
α =[tex]\frac{(0.83*770)^2}{160}[/tex] = 2553 MPa
b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]
b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex] = -0.2005
Next :
calculating the fatigue strength using the relation
Sf = αN^b
N = number of cycles
Sf = 2553 ( 10^4) ^ -0.2005
Sf = 403 MPa
Calculate the maximum moment of the beam
M = 2000 * 0.6 = 1200 N-m
calculating the maximum stress in the beam
∝ = ∝max = [tex]\frac{Mc}{I}[/tex]
Where c = b/2 and I = b(b^3) / 12
hence ∝max = [tex]\frac{6M}{b^3}[/tex] = 6(1200) / b^3 = 7200/ b^3 Pa
note: b is in (meters)
The expression for the factor of safety is written as
n = [tex]\frac{Sf}{\alpha max }[/tex]
Sf = 403, n = 1.5 and ∝max = 7200 / b^3
= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex] making b subject of the formula in other to get the value of b
b = 0.0299 m * 10^3 mm/m
b = 29.9 mm therefore b ≈ 30 mm
since the size factor assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2
the dimension = 30 mm by 30 mm
A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet.
(a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?
Answer:
Explanation:
a ) for transformer which steps down voltage , if V₁ and V₂ be voltage of primary and secondary coil and n₁ and n₂ be the no of turns of wire in them
V₁ /V₂ = n₁ / n₂
Here V₁ = 220 V , V₂ = 5V , n₁ = 2200 n₂ = ?
220 /5 = 2200 / n₂
n₂ = 2200 x 5 / 220
= 50
b )
for 100 % efficiency
input power = output power
V₁ I₁ = V₂I₂
I₁ and I₂ are current in primary and secondary coil
220 x I₁ = 5 x .5
I₁ = .01136 A .
c )
If n₂ = 100
V₁ /V₂ = n₁ / n₂
220 / V₂ = 2200 / 100
V₂ = 10 V
V₁ I₁ = V₂I₂
220 x .01136 = 10 I₂
I₂ = .25 A.
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P
Answer:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Explanation:
The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"
Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:
[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]
Where:
[tex]n[/tex] - Safety factor, dimensionless.
[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])
[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]
[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]
[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]
Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:
[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]
Where:
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
[tex]V[/tex] - Shear force, measured in kilonewtons.
[tex]A[/tex] - Cross section area, measured in square meters.
As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:
[tex]V = \frac{P}{5}[/tex]
[tex]V = \frac{450\,kN}{5}[/tex]
[tex]V = 90\,kN[/tex]
The minimum allowable cross section area is cleared in the shearing stress equation:
[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]
If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:
[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]
[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]
The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:
[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]
The diameter is now cleared and computed:
[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]
[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]
[tex]D = 0.0457\,m[/tex]
[tex]D = 45.7\,mm[/tex]
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
We have that the minimum allowable bolt diameter is mathematically given as
d = 26.65mmFrom the question we are told
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of Assuming P to be P = 425 kN.DiameterGenerally the equation for the stress is mathematically given as
[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]
Therefore
Force = Stress * area
Force = P/2
F= 425,000 N / 2 = 212,500 N
Hence area of each bolt is given as
212,500 = 76.190*( 5* area of each bolt)
area of each bolt = 557.815
Since
area of each bolt=\pi*d^2/4
\pi*d^2/4 = 557.815
d = 26.65mmFor more information on diameter visit
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Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have four arithmetic operations which depend on the values of the selection variables as shown below. V1=0011, V0=0101 and output functions are as follows;
1- A+9's complement of B
2- A+B
3- A+10's complement of B
4- A+1 (add 1 to A)
(You can see question number 3 in the attached file)
Talc and graphite are two of the lowest minerals on the hardness scale. They are also described by terms like greasy or soapy. Both have a crystal structure characterized by sheet-structures at the atomic level, yet they don't behave like micas. What accounts for their unusual physical properties
Answer:
The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.
Explanation:
Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.
Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.
While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.
So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 8.7 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
Answer:
[tex] T_A_C = 6.296 kN [/tex]
[tex] R = 10.06 kN [/tex]
Explanation:
Given:
[tex] T_A_B = 8.7 kN[/tex]
Required:
Find the tension TAC and magnitude R of this downward force.
First calculate [tex] \alpha, \beta, \gamma [/tex]
[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]
[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]
[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]
To Find tension in AC and magnitude R, use sine rule.
[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]
Substitute values:
[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]
Solve for T_A_C:
[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]
[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]
Solve for R.
[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]
[tex] R = 8.7 * 1.156 [/tex]
R = 10.06 kN
Tension AC = 6.296kN
Magnitude,R = 10.06 kN
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickness is 4 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
Explanation:
Given that,
The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]
The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]
The area of the styrofoam [tex]A_s=11m^2[/tex]
The thickness of the styrofoam [tex]t_s=0.20m[/tex]
The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]
The thermal conductivity of the styrofoam [tex]k_s=0.010J(s.m.C^o)[/tex]
Inside and outside temperature difference is ΔT
The heat loss due to conduction in the window is
[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]
The heat loss due to conduction in the wall is
[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]
The net heat loss of the wall and the window is
[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]
The percentage of heat lost by the window is
[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]
You are standing at the edge of the roof of the engineering building, which is H meters high. You see Professor Murthy, who is h meters tall, jogging towards the building at a speed of v meters/second. You are holding an egg and want to release it so that it hits Prof Murthy squarely on top of his head. What formulas describes the distance from the building that Prof Murthy must be when you release the egg?
Answer:
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
Explanation:
First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:
Vf = Vi + gt
where,
Vf = Velocity of egg at the time of hitting head of the Professor
Vi = initial velocity of egg = 0 m/s (Since, egg is initially at rest)
g = acceleration due to gravity
t = time taken by egg to come down
Therefore,
Vf = 0 + gt
t = Vf/g ----- equation (1)
Now, we use 3rd euation of motion for Vf:
2gs = Vf² - Vi²
where,
s = height dropped = H - h
Therefore,
2g(H - h) = Vf²
Vf = √[2g(H - h)]
Therefore, equation (1) becomes:
t = √[2g(H - h)]/g
t = √[2(H - h)/g]
Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:
s = vt
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
Which statements describe how the Fed responds to high inflation? Check all that apply.
It charges banks more interest.
It pays banks less interest.
It sells more securities.
It decreases the money supply.
It increases the money supply.
Answer:
Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply
Explanation:
hope this help edge 21
A shell-and tube heat exchanger (two shells, four tube passes) is used to heat 10,000 kg/h of pressurized water from 35 to 120 oC with 5000 kg/h pressurized water entering the exchanger at 300 oC. If the overall heat transfer coefficient is 1500 W/m^2-K, determine the required heat exchanger area.
Answer:
4.75m^2
Explanation:
Given:-
- Temperature of hot fluid at inlet: [tex]T_h_i = 300[/tex] °C
- Temperature of cold fluid at outlet: [tex]T_c_o = 120[/tex] °C
- Temperature of cold fluid at inlet: [tex]T_c_i = 35[/tex] °C
- The overall heat transfer coefficient: U = 1500 W / m^2 K
- The flow rate of cold fluid: m_c = 10,00 kg/ h
- The flow rate of hot fluid: m_h = 5,000 kg/h
Solution:-
- We will evaluate water properties at median temperatures of each fluid using table A-4.
Cold fluid: Tci = 35°C , Tco = 35°C
Tcm = 77.5 °C ≈ 350 K --- > [tex]C_p_c = 4195 \frac{J}{kg.K}[/tex]
Hot fluid: Thi = 300°C , Tho = 150°C ( assumed )
Thm = 225 °C ≈ 500 K --- > [tex]C_p_h = 4660 \frac{J}{kg.K}[/tex]
- We will use logarithmic - mean temperature rate equation as follows:
[tex]A_s = \frac{q}{U*dT_l_m}[/tex]
Where,
A_s : The surface area of heat exchange
ΔT_lm: the logarithmic differential mean temperature
q: The rate of heat transfer
- Apply the energy balance on cold fluid as follows:
[tex]q = m_c * C_p_c * ( T_c_o - T_c_i )\\\\q = \frac{10,000}{3600} * 4195 * ( 120 - 35 )\\\\q = 9.905*10^5 W[/tex]
- Similarly, apply the heat balance on hot fluid and evaluate the outlet temperature ( Tho ) :
[tex]T_h_o = T_h_i - \frac{q}{m_h * C_p_h} \\\\T_h_o = 300 - \frac{9.905*10^5}{\frac{5000}{3600} * 4660} \\\\T_h_o = 147 C[/tex]
- We will use the experimental results of counter flow ( unmixed - unmixed ) plotted as figure ( Fig . 11.11 ) of the " The fundamentals to heat transfer" and determine the value of ( P , R , F ).
- So the relations from the figure 11.11 are:
[tex]P = \frac{T_c_o - T_c_i}{T_h_i - T_c_i} \\\\P = \frac{120 - 35}{300 - 35} \\\\P = 0.32[/tex]
[tex]R = \frac{T_h_i - T_h_o}{T_c_o - T_c_i} \\\\R = \frac{300 - 147}{120 - 35} \\\\R = 1.8[/tex]
Therefore, P = 0.32 , R = 1.8 ---- > F ≈ 0.97
- The log-mean temperature ( ΔT_lm - cf ) for counter-flow heat exchange can be determined from the relation:
[tex]dT_l_m = \frac{( T_h_i - T_c_o ) - ( T_h_o - T_c_i ) }{Ln ( \frac{( T_h_i - T_c_o )}{( T_h_o - T_c_i )} ) } \\\\dT_l_m = \frac{( 300 - 120 ) - ( 147 - 35 ) }{Ln ( \frac{( 300-120 )}{( 147-35)} ) } \\\\dT_l_m = 143.3 K[/tex]
- The log - mean differential temperature for counter flow is multiplied by the factor of ( F ) to get the standardized value of log - mean differential temperature:
[tex]dT_l = F*dT_l_m = 0.97*143.3 = 139 K[/tex]
- The required heat exchange area ( A_s ) can now be calculated:
[tex]A_s = \frac{9.905*10^5 }{1500*139} \\\\A_s = 4.75 m^2[/tex]
Consider a series RC circuit at the left where C = 6 µ F, R = 2 MΩ, and ε = 20 V. You close the switch at t = 0. Find (a) the time constant for the circuit, (b) the half-life of the circuit, (c) the current at t = 0, (d) the voltage across the capacitor at t = 0, and (e) the voltage across the resistor after a very long time.
Answer:
(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0
Explanation:
Solution
Given that:
C = 6 µ which is = 6 * 10^ ⁻6
R = 2 MΩ, which is = 2 * 10^ 6
ε = 20 V
(a) When it is at the time constant we have the following:
λ = CR
= 6 * 10^ ⁻6 * 2 * 10^ 6
λ =12 seconds
(b) We solve for the half life of the circuit which is given below:
d₀ = d₀ [ 1- e ^ ⁺t/CR
d = decay mode]
d₀/2 = d₀ 1- e ^ ⁺t/12
2^⁻1 = e ^ ⁺t/12
Thus
t/12 ln 2
t = 12 * ln 2
t = 12 * 0.693
t = 8.31 seconds
(c) We find the current at t = 0
So,
I = d₀/dt
I = d₀/dt e ^ ⁺t/CR
= CE/CR e ^ ⁺t/CR
E/R e ^ ⁺t/CR
Thus,
at t = 0
I E/R = 20/ 2 * 10^ 6
= 10µ A
(d) We find the voltage across the capacitor at t = 0 which is shown below:
V = IR
= 10 * 10^ ⁻6 * 2 * 10^ 6
V = 20 V
(e) We solve for he voltage across the resistor.
At t = 0
I = 0
V =0
When you do a vehicle check, what do you NOT need to keep an eye on?
A. Proper tire inflation
B. Cleanliness of windows and mirrors
C. Functioning indicator lights and headlights
D. Blindspot locations
Answer:
Blindspot Location
Explanation:
Just took the quiz
When you do a vehicle check, you do NOT need to keep an eye on Blind spot locations. The correct option is D.
What is Blind spot location?A blind spot is the area of the road that can't be seen by looking forward through windscreen, or by rear-view and side-view mirrors.
While doing vehicle check, we need to check tire inflation, cleanliness of windows and mirrors along with the functioning indicator lights and headlights.
Blind spot locations does not need to be checked.
Thus, the correct option is D.
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A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical shape. Initially, the balloon is filled up with air at the pressure and temperature of 100 kPa and 27°C respectively and the initial diameter (D) of the balloon is 10 m. Then the balloon is heated up to the point that the volume is 1.2 times greater than the original volume (V2 =1.2V1 ). Due to elastic material used in this balloon, the inside pressure ( P ) is proportional to balloonâs diameter, i.e. P = ð¼D, where ð¼ is a constant.
Required:
a. Show that the process is polytropic (i.e. PV" = Constant) and find the exponent n and the constant.
b. Find the temperature at the end of the process by assuming air to be ideal gas.
c. Find the total amount of work that is done by the balloon's boundaries and the fraction of this work that is done on the surrounding atmospheric air at the pressure of 100 kPa.
Answer:
a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]
[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]
We have;
[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]
[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]
[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]
p₂ = 100000/0.941 = 106.265 kPa
[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
4. In its natural state, a soil weighs 2800 lb/cy, while in the loose and compacted states, it weighs 2500 lb/cy and 3300 lb/cy, respectively. a. Find the load and shrinkage factors for this soil. b. How many trucks loads with a capacity of 5 lcy/truck would be required to haul 750,000 ccy of this soil to a project
Answer:
a. load factor = 0.893
shrinkage factor = 0.848
b. Number of Trucks loads = 113,585 Trucks loads
Explanation:
Here, we start by identifying the factors as given in the question.
γn = 2800 lb/cy
γloose = 2500 lb/cy
and γcompacted = 3300 lb/cy
a. Mathematically,
Load factor = γloose/γn = 2500/2800 = 0.893
Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848
b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;
ρlcy = 5
Load factor × Shrinkage factor = ρloose/γn × γn/γcompacted = ρlcy/ρccy
0.893 × 0.848 = 5/ρccy
ρccy =5/(0.893 × 0.848) = 6.603
The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads
Sometimes, steel studs may not be used on outside walls because they are?
Answer:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Explanation:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Effluents from metal-finishing plants have the potential of discharging undesirable quantities of metals, such as cadmium, nickel, lead, manganese, and chromium, in forms that are detrimental to water and air quality. A local metal-finishing plant has identified a wastewater stream that contains 5.15 wt% chromium (Cr) and devised the following approach to lowering risk and recovering the valuable metal. The wastewater stream is fed to a treatment unit that removes 95% of the chromium in the feed and recycles it to the plant. The residual liquid stream leaving the treatment unit is sent to a waste lagoon. The treatment unit has a maximum capacity of 4500 kg wastewater/h. If wastewater leaves the finishing plant at a rate higher than the capacity of the treatment unit, the excess (anything above 4500 kg/h) bypasses the unit and combines with the residual liquid leaving the unit, and the combined stream goes to the waste lagoon.
(a) Without assuming a basis of calculation, draw and label a flowchart of the process. (b) Waste water leaves the finishing plant at a rate m_ 1 ? 6000 kg/h. Calculate the flow rate of liquid to
the waste lagoon, m_ 6?kg/h?, and the mass fraction of Cr in this liquid, x6(kg Cr/kg). (c) Calculate the flow rate of the liquid to the waste lagoon and the mass fraction of Cr in this liquid for m_1 varying from 1000 kg/h to 10,000 kg/h in 1000 kg/h increments. Generate a plot of x6 versus m_ 1 .
(Suggestion: Use a spreadsheet for these calculations.) (d) The company has hired you as a consultant to help them determine whether or not to add capacity to the treatment unit to increase the recovery of chromium. What would you need to know to make this determination? (e) What concerns might need to be addressed regarding the waste lagoon?
Answer:
Explanation:
The solution of all the four parts is provided in the attached figures
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The water and its container are heated to 70oC. Assuming no evaporation, what then will be the depth of the water column if the coefficient of thermal expansion for the glass is 3.8*10-6 mm/mm peroC ?
Answer:
[tex]\mathbf{h_2 =1021.9 \ mm}[/tex]
Explanation:
Given that :
The initial volume of water [tex]V_1[/tex] = 1000.00 mL = 1000000 mm³
The initial temperature of the water [tex]T_1[/tex] = 10° C
The height of the water column h = 1000.00 mm
The final temperature of the water [tex]T_2[/tex] = 70° C
The coefficient of thermal expansion for the glass is ∝ = [tex]3.8*10^{-6 } mm/mm \ per ^oC[/tex]
The objective is to determine the the depth of the water column
In order to do that we will need to determine the volume of the water.
We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:
At temperature [tex]T_1 = 10 ^ 0C[/tex] the density of the water is [tex]\rho = 999.7 \ kg/m^3[/tex]
At temperature [tex]T_2 = 70^0 C[/tex] the density of the water is [tex]\rho = 977.8 \ kg/m^3[/tex]
The mass of the water is [tex]\rho V = \rho _1 V_1 = \rho _2 V_2[/tex]
Thus; we can say [tex]\rho _1 V_1 = \rho _2 V_2[/tex];
⇒ [tex]999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2[/tex]
[tex]V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }[/tex]
[tex]V_2 = 1022.40 \ mL[/tex]
[tex]v_2 = 1022400 \ mm^3[/tex]
Thus, the volume of the water after heating to a required temperature of [tex]70^0C[/tex] is 1022400 mm³
However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:
[tex]V_1 = A_1 *h_1[/tex]
The area of the water before heating is:
[tex]A_1 = \dfrac{V_1}{h_1}[/tex]
[tex]A_1 = \dfrac{1000000}{1000}[/tex]
[tex]A_1 = 1000 \ mm^2[/tex]
The area of the heated water is :
[tex]A_2 = A_1 (1 + \Delta t \alpha )^2[/tex]
[tex]A_2 = A_1 (1 + (T_2-T_1) \alpha )^2[/tex]
[tex]A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2[/tex]
[tex]A_2 = 1000.5 \ mm^2[/tex]
Finally, the depth of the heated hot water is:
[tex]h_2 = \dfrac{V_2}{A_2}[/tex]
[tex]h_2 = \dfrac{1022400}{1000.5}[/tex]
[tex]\mathbf{h_2 =1021.9 \ mm}[/tex]
Hence the depth of the heated hot water is [tex]\mathbf{h_2 =1021.9 \ mm}[/tex]
Caulking is recommended around the edges of partitions between apartments to... Group of answer choices reduce the need for trim. reduce sound transmission. reduce heat loss. increase the fire rating of the partition
Answer:
Reduce sound transmission.
Explanation:
A caulking is a flexible material used to seal joints, cracks or gaps formed between building materials and pipes against leakage.
Caulking is recommended around the edges of partitions between apartments to reduce sound transmission.
Hence, in the event that an individual notices that air or sound is gaining entrance into their apartment, a caulking can be used to mitigate this noise or unwanted sound.
The caulking when applied to the gap or edges of partitions between apartments would create a tight seal and block the flow or entry of air, thereby reducing sound transmission.
The lower half of a 7-m-high cylindrical container is filled with water (rho = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and the bottom of the cylinder. (Round the final answer to one decimal place.)
Answer:
Pressure difference (ΔP) = 63,519.75 kpa
Explanation:
Given:
ρ = 1,000 kg/m³
Height of cylindrical container used (h) = 7m / 2 = 3.5m
Specific gravity (sg) = 0.85
Find:
Pressure difference (ΔP).
Computation:
⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ] ∵ [ g = 9.81]
⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [1,850]
⇒ Pressure difference (ΔP) = 63,519.75 kpa
