Answer:
Explanation:
A negative binary number is represeneted by its 2's complement value. To get 2's complement, you just need to invert the bits and add 1 to it. So the formula is:
twos_complement = ~val + 1
So you start out with 22 and you want to make it negative.
22₁₀ = 0001 0110₂
~22₁₀ = 1110 1001₂ inverting the bits
~22₁₀ + 1 = 1110 1010₂ adding 1 to it.
so -22₁₀ == ~22₁₀ + 1 == 1110 1010₂
Do the same process for 16-bits and 32-bits and you'll find that the most significant bits will be padded with 1's.
-22₁₀ = 1110 1010₂ 8-bits
-22₁₀ = 1111 1111 1110 1010₂ 16-bits
-22₁₀ = 1111 1111 1111 1111 1111 1111 1110 1010₂ 32-bits
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]
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:
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
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]
: Explain why testing can only detect the presence of errors, not their absence?
Answer:
The goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.
Explanation:
Increase the sampling time by a factor of 10 (to 0.1 seconds), keeping the frequency of the square wave the same, and observe the delay. Discuss relationship between sampling time and delay from one board to another.
Answer:
Time delay increases
Explanation:
Time delay is the delay between occurance of signal. If sampling time that is time between two samples is increased, the delay in the occurance of regenerated samples is also increased.
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.
g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70% of the carnot effieciency. if 2kj of work are produced, how much heat is rejected to the low temperature reservior
Answer:
Heat rejected to cold body = 3.81 kJ
Explanation:
Temperature of hot thermal reservoir Th = 1600 K
Temperature of cold thermal reservoir Tc = 400 K
efficiency of the Carnot's engine = 1 - [tex]\frac{Tc}{Th}[/tex]
eff. of the Carnot's engine = 1 - [tex]\frac{400}{600}[/tex]
eff = 1 - 0.25 = 0.75
efficiency of the heat engine = 70% of 0.75 = 0.525
work done by heat engine = 2 kJ
eff. of heat engine is gotten as = W/Q
where W = work done by heat engine
Q = heat rejected by heat engine to lower temperature reservoir
from the equation, we can derive that
heat rejected Q = W/eff = 2/0.525 = 3.81 kJ
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
two opposite poles repel each other
Answer:
South Pole and South Pole or North Pole and North Pole.
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]
An aluminium bar 600mm long with a diameter 40mm has a hole drilled in the centre of which 30mm in diameter and 100mm long if the modulus of elasticity is 85GN/M2 calculate the total contraction oon the bar due to comprehensive load of 160KN.
Answer:
Total contraction on the bar = 1.238 mm
Explanation:
Modulus of Elasticity, E = 85 GN/m²
Diameter of the aluminium bar, [tex]d_{Al} = 40 mm = 0.04 m[/tex]
Load, P = 160 kN
Cross sectional area of the aluminium bar without hole:
[tex]A_1 = \frac{\pi d_{Al}^2 }{4} \\A_1 = \frac{\pi 0.04^2 }{4}\\A_1 = 0.00126 m^2[/tex]
Diameter of hole, [tex]d_h = 30 mm = 0.03 m[/tex]
Cross sectional area of the aluminium bar with hole:
[tex]A_2 = \frac{\pi( d_{Al}^2 - d_{h}^2)}{4} \\A_2 = \frac{\pi (0.04^2 - 0.03^2) }{4}\\A_2 = 0.00055 m^2[/tex]
Length of the aluminium bar, [tex]L_{Al} = 600 mm = 0.6 m[/tex]
Length of the hole, [tex]L_h = 100mm = 0.1 m[/tex]
Contraction in aluminium bar without hole [tex]= \frac{P * L_{Al}}{A_1 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.6}{0.00126 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 96000/107100000
Contraction in aluminium bar without hole = 0.000896
Contraction in aluminium bar with hole [tex]= \frac{P * L_{h}}{A_2 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.1}{0.00055 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 16000/46750000
Contraction in aluminium bar without hole = 0.000342
Total contraction = 0.000896 + 0.000342
Total contraction = 0.001238 m = 1.238 mm
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
The drum has a mass of 50 kg and a radius of gyration about the pin at O of 0.23 o k m = . If the 15kg block is moving downward at 3 / m s , and a force of P N =100 is applied to the brake arm, determine how far the block descends from the instant the brake is applied until it stops. Neglect the thickness of the handle. The coefficient of kinetic friction at the brake pad is 0.5 k = .
Note: The diagram referred to in this question is attached as a file below.
Answer:
The block descended a distance of 9.75m from the instant the brake is applied until it stops.
Explanation:
For clarity and easiness of expression, the calculations and the Free Body Diagram are contained in the attached file. Check the attached file below.
The block descended a distance of 9.75 m
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.
Participating in extracurricular activities in high school helps:
Answer:
Develop social skills
Explanation:
Answer:
strengthen your college applications
Explanation:
When an electrical signal travels through a conductive wire, it produces an electromagnetic (EM) field. Likewise, when an EM field encounters a conductive wire, it produces a proportional electrical current.
A. True
B. False
Answer:
A. True
Explanation:
When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.
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.
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 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]
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]
An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 750 kPa and 48°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank. (Round the final answers to the nearest whole number.)
Answer:
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
Explanation:
An ideal gas is represented by the following model:
[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]
Where:
[tex]P[/tex] - Pressure, measured in kilopascals.
[tex]V[/tex] - Volume, measured in cubic meters.
[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.
[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.
[tex]T[/tex] - Temperature, measured in Kelvin.
[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]
As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:
[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]
Now, final pressure is cleared:
[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]
[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]
[tex]P_{2} = 375\,kPa[/tex]
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
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
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°
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
Consider a classroom for 56 students and one instructor, each generating heat at a rate of 100 W. Lighting is provided by 18 fluorescent lightbulbs, 40 W each, and the ballasts consume an additional 10 percent. Determine the rate of internal heat generation in this classroom when it is fully occupied. The rate of internal heat generation in this classroom when it is fully occupied is W.
Answer:
What is the probability of selecting the 4 of spade or black diamond from a deck of 52 playing cards?
a) 2/52
b) 4/52
c) 3/52
d) 1/5
Explanation:
there is usually a positive side and a negative side to each new technological improvement?
Answer:
positive sides:
low cost improves production speedless timeeducational improvementsnegative sides:
unemployment lot of space required increased pollution creates lots of ethical issuesFive 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
https://brainly.com/question/8552546
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
