Where are your answer options?
Answer:
Implement a hazard communication program
Explanation: i took the quiz
An array of eight aluminum alloy long fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 340 K and the ambient air is at 300 K, how much power do they dissipate if the combined convection and radiation heat transient coefficient is estimated to be 8 W/m2K? The alloy has a conductivity of 175 W/mKand the heat transfer from the tip is negligible.
Answer:
0.08704 W
Explanation:
converting the mm to m (1000mm = 1m)
cross-sectional area of the fins, Ac = (0.003) (0.0004) = 0.0000012m^2
The wetted perimeter of the cross-section, P = 2 (0.003 + 0.0004) = 0.0068m
Thickness of solid in direction of heat flow, B^2 = (heat transient coefficient, h) (The wetted perimeter of the cross-section, P) ÷ (Thermal conductivity, k) (cross-sectional area of the fins, Ac)
B^2 = (8 W/m2K)(0.0068m) ÷ (175 W/mK)(0.0000012m^2)
=259.0476m^-2
B= square root of the result
B = 16.09m^-1
we now look for:
The Coordinate, x = B, multiplied by Length, L
x = (16.09m^-1) (0.04m) = 0.6436
finding the side area of a fin = P multiplied by Length, L
= 0.0068m X 0.04m = 0.000272m^2
Neglecting inefficiency, assuming the fins are all 100% efficient, the power they would dissipate =
h, Heat-transfer coefficient (PL) (temperature of at the base - temperature at the ambient air)
= (8) (0.000272m^2)(340 K- 300k)
= 0.08704 W
what's nested piezometer?
Explanation:
Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube
For a steel alloy it has been determined that a carburizing heat treatment of 9-h duration will raise the carbon concentration to 0.38 wt% at a point 1.2 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm position for an identical steel and at the same carburizing temperature.
Answer:
t2 = 256 hours
Explanation:
Given data:
Carbon concentration ( C ) at 1.2mm from surface, C = 0.38 wt%
Duration( t ) of heat treatment for 0.38wt% at 1.2mm = 9-hr
Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm
Assuming same concentration of 0.38wt% we will apply Fick's second law for constant surface concentration
attached below is the remaining part of the solution
x1 = 1.2 mm
x2 = 6.4 mm
t1 = 9-hr
t2 = ?
t2 = [tex](\frac{6.4}{1.2} )^{2} * 9[/tex] = 256 hours
1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel is 200 GPa and 75 GPa for aluminium. The allowable normal stress in steel is 220 MPa and 100 MPa in aluminium. Determine the largest permissible bending moment when the composite bar is bent about horizontal axis. a
Answer:
1.933 KN-M
Explanation:
Determine the largest permissible bending moment when the composite bar is bent horizontally
Given data :
modulus of elasticity of steel = 200 GPa
modulus of elasticity of aluminum = 75 GPa
Allowable stress for steel = 220 MPa
Allowable stress for Aluminum = 100 MPa
a = 10 mm
First step
determine moment of resistance when steel reaches its max permissible stress
next : determine moment of resistance when Aluminum reaches its max permissible stress
Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M
attached below is a detailed solution
The two structural members, one of which is in tension and the other in compression, exert the indicated forces on joint o. determine the magnitude r of the resultant r of the two forces and the angle θ which r makes with the positive x axis (measured counterclockwise from the x axis).
Discuss the nature of materials causing turbidity in
(a) River water during flash flood
(b) Polluted river water
(c) Domestic wastewater
Answer:
a
Explanation:
Construct the plane-stress yield envelopes in a principle stress space for both the Tresca and the von Mises yield theories using your calculated value of the yield strength to scale the envelopes. Indicate the two equivalent load paths corresponding to pure shear on the yield envelopes. Calculate the shear yield strength of Al 6061-T6 aluminum predicted by the above theories.
Answer:
Explanation:
The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.
From the given information:
The elastic modulus is:
[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]
Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.
The shear yield strength by using von Mises criteria is estimated as;
[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]
The shear yield strength by using Tresca criteria is:
[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]
electrical engineering
Answer:
Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.
Explanation:
what’s your favorite color of the alphabet? if so, what fruit is it?
Hi sorry I need points I'm New
Answer: My favorite color is Red favorite letter C favorite fruit watermelon.
Forces always act in equal and opposite pairs
Consider the following chain-reaction mechanism for the high-temperatureformation of nitric oxide, i.e., the Zeldovich mechanism:
O+ N2 → NO + N
N + O2 → NO+ O
Write out expressions for d[NO] / dt and d[N] / dt.
Answer: hello attached below is the properly written chain reaction to your question
answer :
d[NO] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]
d[N] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]
Explanation:
write out expressions for d[NO] / dt and d[N] / dt
Given :
properly written chain reaction ( attached below)
Expression for d[NO] / dt can be written as
[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]
Expression for d[N] / dt can be written as
[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]
In a metal-oxide-semiconductor (MOS) device, a thin layer of SiO2 (density = 2.20 Mg/m3) is grown on a single crystal chip of silicon. How many Si atoms and how many O atoms are present per square millimeter of the oxide layer? Assume that the layer thickness is 160 nm.
Answer:
3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O
Explanation:
It is all about unit conversions.
The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.
160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm
By the way, now we can determine the volume of MOS, in order to work with density.
1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³
But density is mg/m³, so we convert mm³ to m³
1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³
Now, we apply density to determine the mass of MOS
Density = mass /volume → Density . volume = mass
1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg
To make more easier the calculate, we convert mg to g.
3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g
To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol
We need to know moles of Si and O₂ in the MOS
Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles
1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:
5.86×10⁻¹⁸ mol of SiO₂ may have:
(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si
(5.86×10⁻¹⁸ .2) /1 = 1.17×10⁻¹⁷ moles of O₂
Let's count the atoms (1 mol of anything contain NA particles)
5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si
1.17×10⁻¹⁷ mol of O₂ . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O
The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O
What is the number of atoms present?We are given;
Area of MOS device; A = 1 mm²
Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm
Formula for volume is;
V = Area * thickness
V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³
Converting volume to m³ gives;
V = 1.6 × 10⁻¹³ m³
Now, to get the mass of MOS, we will use the formula;
Mass = Density * volume
we are given density = 2.20mg/m³
Thus;
Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³
Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g
From periodic table, the molar mass of SiO₂ = 60.08 g/mol
Then;
Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol
Number of moles of MOS = 5.86 × 10⁻¹⁸ moles
Now, 1 mol of SiO₂ is formed from 1 mol of Si and 2 mol of O. Thus;
5.86×10⁻¹⁸ mol of SiO₂ will have;
5.86×10⁻¹⁸ moles of Si
and (5.86×10⁻¹⁸ .2) = 11.72 × 10⁻¹⁸ moles of O₂
From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;
5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si
11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O
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The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and usingit topreheat the water entering the boiler. In this process,liquid water at 50oC and 1000 kPa is mixed withsuperheated steam at 200oC and 1000 kPa. If the plantoperators want to produce a saturated liquid at 1000kPa, what ratio of mass flow rates of water andsuperheatedsteam are required
Answer:
Explanation:
This is Answer....
When using the superposition theorem, each ideal voltage and current source must be replaced by an open circuit.
true
false
The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter of the rod be not to deform
Answer:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Determine (a) the maximum flow rate of water (5-point) and (b) the pressure difference across the pump (5-point). Assume the elevation difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible
Answer:
a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s
b) The pressure difference across the pump is approximately 293.118 kPa
Explanation:
The efficiency of the pump = 78%
The power of the pump = 5 -kW
The height of the pool above the underground water, h = 30 m
The diameter of the pipe on the intake side = 7 cm
The diameter of the pipe on the discharge side = 5 cm
a) The maximum flowrate of the pump is given as follows;
[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]
Where;
P = The power of the pump
Q = The flowrate of the pump
ρ = The density of the fluid = 997 kg/m³
h = The head of the pump = 30 m
g = The acceleration due to gravity ≈ 9.8 m/s²
[tex]\eta_t[/tex] = The efficiency of the pump = 78%
[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]
[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s
The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s
b) The pressure difference across the pump, ΔP = ρ·g·h
∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa
The pressure difference across the pump, ΔP ≈ 293.118 kPa
A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter
Answer:
shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm
Explanation:
Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6
Shear stress of material ( Z ) ≤ 500 KPa
power transmitted by shaft ( P ) = 1.5MW of mechanical power
Revolution ( N ) = 1500 rev/min
Calculate shaft Diameter
Given that: P = [tex]\frac{2\pi NT}{60}[/tex] ---- 1
therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 ) = 9.554 KN-M
next
[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]
hence ; T = Z[tex]_{p} *Z[/tex]
attached below is the remaining part of the solution
Automotive gas turbines have been under development for decades but have not been commonly used in automobiles. Yet helicopters routinely use gas turbines. Explore why different types of engines are used in these respective applications. Compare selection factors such as performance, power-to-weight ratio, space requirements, fuel availability, and environmental impact.
Required:
Summarize your findings in a report with at least three references.
Answer:
Gas turbines in Helicopters require lesser space.
Explanation:
[1] In terms of Space Requirements:
The gas used in helicopters requires lesser space as compared to Automotive gas turbines. The gas in automobile have higher thermal efficiency.
[2]. In terms of Environmental impact:
The occurrence of environmental solution is very slim when used in helicopters' engines.
[3]. In terms of power-to-weight ratio:
The vibrations in engines of helicopters make it to have lesser efficiency as compared to automobile.
[4]. In terms of Fuel availability:
Fuel is available. Automobile can make use of gas as fuel.
Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0) = 0 and derive y"
Answer:
Explanation:
Given that:
[tex]y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)[/tex]
where;
[tex]g'(0) = \dfrac{1}{m}[/tex] and [tex]mg"+kg = 0[/tex]
[tex]\text{Using Leibniz Formula to prove the above equation:}[/tex]
[tex]\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt[/tex]
So, [tex]y = \int ^t_0 g' (t-s) f(s) \ ds[/tex]
[tex]\text{By differentiation with respect to t;}[/tex]
[tex]y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds[/tex]
[tex]y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)[/tex]
[tex]Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\[/tex]
[tex]Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}[/tex]
A switch that can open or close an electric circuit can be used to?
Answer:
When a switch is in the "off" position the circuit is open. Electric charges cannot flow when a switch is in the off position.
Explanation:
A switch that can open or close an electric circuit can be used to stop the flow of current.
What is a switch?A switch can be defined as an electrical component (device) that is typically designed and developed for interrupting the flow of current or electrons in an electric circuit.
This ultimately implies that, a switch that can open or close an electric circuit can be used to stop the flow of current.
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Although many countries have issues with soil erosion due to deforestation, some of the most serious effects are seen
in which country?
Derive the next state equations for each type (D, T, SR, and JK) of basic memory element. The next state equation is a symbolic equation describing the next state (Q ) as a function of the inputs (D,T,SR, or JK) and state (Q). In order to determine the next state equations for a a JK memory element, build a 3-variable Kmap with Q, J, and K as the inputs. The entries in the Kmap should be Q . Solving this Kmap will yield the next state equation. Show all work for full credit.
Answer:
Attached below is the derived next state equations
Explanation:
Attached below is the derived next state equations
used for the solution of the given problem.
someone should help me and explain stress strain curve
Answer:
In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).
Explanation:
please mark brainliest
How does distribution add value to goods and services being sold,
including intellectual property?
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Explanation:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.
Explanation:
hope it helps <33
When choosing a respirator for your job, you must conduct a
test.
A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 22,000 N (4950 lbf), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0
Answer:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
attached below is a detailed solution
the removed soil at an excavation site is also called spoil?
Answer:
True, That is correct. Soil removed from an excavation site is indeed called spoil.
Spoil definition: The waste material (such as soil) brought up during the course of an excavation.
Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D
Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Thus, The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.
It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.
Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.
Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Learn more about Soil, refer to the link:
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Clear Lake has a surface area of 70 ha. In April the inflow of the lake was 1.5 m3/sec. A dam regulated the outflow of the lake to be 1.25 m3/sec. If the precipitation recorded for the month of April was 7.62 cm and the storage volume increased by an estimated 650,000 m3. What is the estimated evaporation in cubic meters?
Answer:
The estimated evaporation is 51340 cubic meters.
Explanation:
Let suppose that precipitation is very small in comparison with depth of the Clear Lake. The monthly change in the volume of the lake ([tex]\Delta V[/tex]), in cubic meters, is estimated by the following formula:
[tex]\Delta V = A\cdot \Delta z + (\dot V_{in}-\dot V_{out})\cdot \Delta t -V_{evap}[/tex] (1)
Where:
[tex]A[/tex] - Surface area of the lake, in square meters.
[tex]\Delta z[/tex] - Water precipitation, in meters.
[tex]\dot V_{in}[/tex] - Average water inflow, in cubic meters per second.
[tex]\dot V_{out}[/tex] - Average water outflow, in cubic meters per second.
[tex]\Delta t[/tex] - Monthly time, in seconds.
[tex]V_{evap}[/tex] - Evaporation, in cubic meters.
If we know that [tex]A = 700000\,m^{2}[/tex], [tex]\Delta z = 0.0762\,m[/tex], [tex]\dot V_{in} = 1.5\,\frac{m^{3}}{s}[/tex], [tex]\dot V_{out} = 1.25\,\frac{m^{3}}{s}[/tex], [tex]\Delta t = 2.592\times 10^{6}\,s[/tex] and [tex]\Delta V = 650000\,m^{3}[/tex], then the estimated evaporation is:
[tex]650000 = 701340-V_{evap}[/tex] (2)
[tex]V_{evap} = 51340\,m^{3}[/tex]
The estimated evaporation is 51340 cubic meters.
The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Answer:
Magnitude of force P = 25715.1517 N
Explanation:
Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.
To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Proof -
Given that,
Diameter = 12 mm = 0.012 m
Length = 0.6 m
[tex]\theta[/tex] = 0.015°
Youngs modulus of elasticity of 34 stainless steel is 193 GPa
Now,
By applying the conditions of equilibrium, we have
∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0
If ∑[tex]M_{A}[/tex] = 0
⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0
⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0
⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]
If ∑[tex]M_{B}[/tex] = 0
⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3
⇒[tex]F_{AD}[/tex] ×3 = P
⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]
Now,
Area, A = [tex]\frac{\pi }{4} X (0.012)^{2}[/tex] = 1.3097 × 10⁻⁴ m²
We know that,
Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]
Now,
[tex]\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 9.1626 × 10⁻⁹ P
[tex]\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 1.83253 × 10⁻⁸ P
Given that,
[tex]\theta[/tex] = 0.015°
⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad
So,
[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]
⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9
⇒P = 25715.1517 N
∴ we get
Magnitude of force P = 25715.1517 N
The magnitude of Force P is; P = 25715.15 N
What is the magnitude of the force?
If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;
Taking moments about point A, we have;
(F_bc * 0.9) - P(0.6) = 0
F_bc = ²/₃P
Taking moments about B gives;
P(0.3) - F_ad * 0.9 = 0
F_ad = ¹/₃P
Normal stress for BC is;
σ_bc = F_bc/A_bc
σ_bc = (²/₃P)/(π * 0.006²)
σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²
σ_ad = (¹/₃P)/(π * 0.006²)
σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²
We know that;
Elongation is; ΔL = PL/AE = (P/A) * (L/E)
Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa
Thus;
ΔL_bc = (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))
ΔL_bc = 1.83253P × 10⁻⁸
Likewise;
ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))
ΔL_ad = 9.1626P × 10⁻⁹ m
Converting the beam tilt angle from degrees to radians gives;
θ = 0.015° = 0.00026179939 rads
Using small angle analysis, we can say that;
θ = (ΔL_bc - ΔL_ad)/36
θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36
Solving gives P = 25715.15 N
Read more about Magnitude of Force at; https://brainly.com/question/13370981
A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.
Answer: hello your question lacks the required diagram attached below is the diagram
answer : 29528.1 N/m^2
Explanation:
Given data :
dimensions of tank :
Length = 5-m
Width = 4-m
Depth = 2.5-m
acceleration of tank = 2m/s^2
Determine the maximum gage pressure in the tank
Pa ( pressure at point A ) = s*g*h1
= 10^3 * 9.81 * 3.01
= 29528.1 N/m^2
attached below is the remaining part of the solution