Answer:
104,576 cycles
Explanation:
Step 1: identify given parameters
Ultimate strength of steel ([tex]S_{ut}[/tex])= 120 Kpsi
stress amplitude ([tex]\alpha_{a}[/tex])= 70 kpsi
life of the specimen (N) = ?
[tex]N = (\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]
where a and b are coefficient of fatigue cycle
Step 2: calculate the the endurance limit of specimen
[tex]S_{e} = 0.5*S_{ut}[/tex]
[tex]S_{e}[/tex] = 0.5*120 = 60 kpsi
Step 3: calculate coefficient 'a'
[tex]a=\frac {(0.8XS_{ut})^2}{S_{e}}[/tex]
[tex]a=\frac {(0.8X120)^2}{60}[/tex]
[tex]a= 153.6 kpsi
Step 4: calculate the coefficient 'b'
[tex]b =-\frac{1}{3}log(\frac{f*S_{ut} }{S_{e}})[/tex]
[tex]b =-\frac{1}{3}log(\frac{0.8*120}{60})[/tex]
[tex]b =-0.0680
Step 5: calculate the life of the specimen
[tex]N=(\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]
[tex]N=(\frac{70}{153.6})^\frac{1}{-0.068}[/tex]
[tex]N=104,576 cycles [/tex]
∴ the life (N) of the steel specimen is 104,576 cycles
An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions. If the concentrations of Ni2+ and Fe2+ ions are 0.002 M and 0.40 M, respectively, what voltage is generated at 25°C? (The respective standard reduction potentials for Ni and Fe are −0.250 V and −0.440 V.) (A) −0.76 V (B) −0.26 V (C) +0.12 V (D) +0.76 V
Answer:C 0.12 V
Explanation:
Given
Concentration of [tex]Fe^{2+} M_1=0.40 M[/tex]
Concentration of [tex]Ni^{2+} M_2=0.002 M[/tex]
Standard Potential for Ni and Fe are [tex]V_2=-0.25 V[/tex] and [tex]V_1=-0.44 V[/tex]
[tex]\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})[/tex]
[tex]\Delta V=-0.25-(-0.44)-\frac{0.0592}{2}\log (\frac{0.4}{0.002})[/tex]
[tex]\Delta V=0.12\ V[/tex]
A rectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. The stilling basin is also 20 feet wide and has a water depth of 8 ft determined from the downstream channel condition. What is the length of the stilling basin? What is the height of the endsill?
Answer:
Length of stilling basin = 32.9 feet
Height of end sill = 6.58 feet
Explanation:
Discharge = Q = 400 ft^3 /sec
Slope = 2.5 ft
Width = 20 feet
n = 0.013
we will assume the depth of flow as "d"
Q = 1/n (R)^2/3 (slope)^1/2 A ( here R is the hydraulic Radius)
by substituting the given data in above formula we get:
400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d
R = A/P
here, A is the flow area and P is the wetted perimeter
400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d
d = 1.42 feet
Depth of stilling channel before the jump will be = d1 = 8 feet
Depth of stilling channel after the jump will be = d2 = 1.42 feet
Length of stilling basin = 5(d2 - d1)
= 5( 8 - 1.42)
Length of stilling basin = 32.9 feet
Now calculating the height of end sill:
Jump height = (8 - 1.42)
Height of end sill = 6.58 feet
The form of mechanical weathering that occurs when a magma chamber starts expanding after the overlying volcano has been removed is known as what?
Answer:
Exfoliation.
Explanation:
Exfoliation is a form of chemical weathering that occurs when sheets of rocks peel from a massive rock's surface. It is also the unloading and sheeting of stress in a rock that expands the magma chamber.
The four-wheel-drive all-terrain vehicle has a mass of 320 kg with center of mass G2. The driver has a mass of 82 kg with center of mass G1.
If all four wheels are observed to spin momentarily as the driver attempts to go forward, what is the forward acceleration of the driver and ATV?
The coefficient of friction between the tires and the ground is 0.59.
The mass of the car plus its occupants is calculated using Newton's second law of motion. After determining the net force by subtracting the force of friction from the force exerted by the wheels, it is divided by the given acceleration to find the mass, which is approximately 1027.78 kg.
Explanation:The question pertains to the application of Newton's laws of motion and forces to determine the mass of a vehicle, given the acceleration, force exerted, and the opposing forces of friction. The calculation is based on Newton's second law, which states that the net force acting on an object is equal to the mass of the object times its acceleration (F = ma). To find the mass of the car plus its occupants, we start with calculating the net force (Fnet) which is the difference between the force the wheels exert backward on the road and the force of friction. By applying Newton's second law (Fnet = m × a), we rearrange it to solve for mass (m = Fnet / a).
Net Force Equation: Fnet = Force exerted - Friction
= 2100 N - 250 N
= 1850 N
Now, we can determine the mass using the net force and the given acceleration.
Mass (m) = Fnet / a
= 1850 N / 1.80 m/s2
= 1027.78 kg
Therefore, the mass of the car plus its occupants is approximately 1027.78 kg.
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype will be operated in seawater. Determine the speed ofthe prototype to ensure Reynolds number similarity‘
Answer:
Explanation:
Given
scale i.e. [tex]L_r=1:15[/tex]
Using Reynolds number similarity
[tex](Re)_m=(Re)_p[/tex]
[tex](\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p[/tex]
Properties of air
[tex]\nu _{air}=1.57\times 10^{-4} ft/s[/tex]
Properties of sea water
[tex]\nu _{sea}=1.26\times 10^{-5} ft/s[/tex]
[tex]\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )[/tex]
[tex]V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )[/tex]
[tex]V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}[/tex]
[tex]V_p=0.963\ ft/s[/tex]
Which of the following is the definition of keyspace? The range of values that construct a cryptosystem key. The process of applying an algorithm to cleartext (or plaintext) data, resulting in a ciphertext. Provides a stronger cryptographic result with a shorter key. Ensuring that only the intended recipient can read the data.
Answer:
A
Explanation:
A key space is a set of all arrangements of a key, when given a fixed key size.
It is the total number of possible values of keys in a cryptographic algorithm or other security measures such as a password.
A key which is a byte long (or 8 bits) would consist of 256 possible keys.
Flue gases from an incinerator are released to atmosphere using a stack that is 0.6 m in diameter and 10.0 m high. The outer surface of the stack is at 40°C and the surrounding air is at 10°C. Determine the rate of heat transfer from the stack, assuming
(a) there is no wind and
(b) the stack is exposed to 20 km/h winds.
To determine the rate of heat transfer from a stack, natural convection and radiation must be considered when there's no wind, and forced convection must be considered when there are 20 km/h winds. Specific calculation details depend on heat transfer coefficients and empirical correlations.
Explanation:To calculate the rate of heat transfer from an incinerator stack, we need to consider the mode of heat transfer. Without wind, the transfer will primarily be through natural convection and radiation. With wind, forced convection becomes significant.
(a) For no wind, we can use equations for natural convection and radiation to determine the heat loss. For convection, Newton's law of cooling \\( ext{Q} = hA\\Delta\\text{T}\\) can be applied where \\('h'\\) is the heat transfer coefficient, \\('A'\\) is the surface area, and \\('\\Delta\\text{T}'\\) is the temperature difference. For radiation, we can use the Stefan-Boltzmann law, considering the emissivity of the stack's material and the temperatures of the surface and the environment.
(b) For 20 km/h winds, the forced convection equation will be used. The heat transfer coefficient \\('h'\\) will be greater due to the wind, increasing the rate of heat transfer. The formula still follows \\( ext{Q} = hA\\Delta\\text{T}\\), but \\('h'\\) will change based on empirical correlations for flow past a cylinder.
Note that specifics of the calculation aren't provided as they require detailed coefficients and calculations beyond the scope of this question.
A) A cross-section of a solid circular rod is subject to a torque of T = 3.5 kNâ‹…m. If the diameter of the rod is D = 5 cm, what is the maximum shear stress? include units.B) The maximum stress in a section of a circular tube subject to a torque is Tmax = 37 MPa . If the inner diameter is Di = 4.5 cm and the outer diameter is Do = 6.5 cm , what is the torque on the section? include units.
Answer:
[tex]\tau_{max} = 142.6[/tex] MPa
T = 1536.8 N m
Explanation:
Given data:
Torque = 3.5 k N m = 3.5*10^3 N.m
Diameter D = 5 cm = 0.05 m
a) from torsional equation we have
[tex]\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}[/tex]
[tex]\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}[/tex]
solving for [tex]\tau_{max}[/tex]
[tex]\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}[/tex]
[tex]\tau_{max} = 142.6[/tex] MPa
B)
[tex]\tau = 37 MPa = 37 \times 10^6[/tex] Pa
[tex]D_i = 4.5 cm = 0.045[/tex] m
[tex]D_o = 6.5 cm = 0.065[/tex] m
[tex]\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}[/tex]
[tex]\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}[/tex]
T = 1536.8 N m
Consider a model of a wing-body shape mounted in a wind tunnel. The flow conditions in the test section are standard sea-level properties with a velocity of 100 m/s. The wing area and chord are 1.5 m2 and 0.45 m, respectively. Using the wind tunnel force and moment-measuring balance, the moment about the center of gravity when the lift is zero is found to be -12.4 N-m. When the model is pitched to another angle of attack, the lift and moment about the center of gravity are measured to be 3675 N and 20.67 N-m, respectively. Calculate the value of the moment coefficient about the aerodynamic center and the location of the aerodynamic center.
Answer:
A. -0.003
B. 0.02
Explanation:
Step 1: identify the given parameters
Giving the following parameters
Wing area (S)= 1.5 m²
Wing chord (C) = 0.45 m
Velocity (V) = 100 m/s
moment about center of gravity(Mcg) = -12.4 N-m
at another angle of attack, L = 3675 N and Mcg = 20.67 N-m
Step 2: calculate the value of the moment coefficient about the aerodynamic center (Cmcg)
[tex]q_{∞} =\frac{1}{2}\rho*v^{2}[/tex]
[tex]q_{∞} =\frac{1}{2}1.225*100^{2}[/tex]= 6125 N/m²
[tex]C_{mcg,w} =\frac{M_{cg,w} }{q_{∞}*S*C }[/tex]
[tex]C_{mcg,w} =\frac{-12.4}{6125*1.5*0.45 }[/tex] = -0.003
[tex]C_{mcg,w}= C_{ac,w}= -0.003[/tex] at zero lift
Step 3: calculate coefficient of lift
Cl = L/q*s
Cl = 3675/6125*1.5 = 0.4
Step 4: calculate the location of the aerodynamic center
New moment coefficient about the aerodynamic center (Cmcg):
[tex]C_{mcg} =\frac{20.67}{6125*1.5*0.45}[/tex] = 0.005
[tex]C_{mcg,w} = C_{ac} ,w + C_{l}(h-h_{ac})[/tex]
[tex]h-h_{ac}= \frac{C_{mcg,w} -C_{ac,w}}{C_{l} }[/tex]
[tex]h-h_{ac}= \frac{0.005-(-0.003)}{0.4}[/tex]=0.02
the location of the aerodynamic center = 0.02
A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B. The surface has a coefficient of friction of 0.3. Determine the value of the horizontal force P necessary to cause motion of the chest to the right, and determine if the motion is sliding or tipping. The value of P is N. The motion is .
The value of P needed to cause motion is 142.5 N, and it's a sliding motion.
Multiply the coefficient of static friction (0.3) by the normal force (475 N) to find the maximum static friction force.
Static friction equals
= [tex]0.3 * 475 N[/tex]
= 142.5 N.
To initiate motion, the applied force (P) must overcome static friction. Thus,
P=142.5 N.
If P was less than s, the chest wouldn't move. Since P equals fs, the motion is sliding. Thus, the value of P needed to cause motion is 142.5 N, and it's a sliding motion.
Percy Motors has a target capital structure of 40% debt and 60% common equity, with no preferred stock. The yield to maturity on the company’s outstanding bonds is 9%, and its tax rate is 40%. Percy’s CFO estimates that the company’s WACC is 9.96%. What is Percy’s cost of common equity?
Answer:
13 %
Explanation:
rd= 9%
T = 40%
WACC = 9.96%
wd = 40%
wc = 60%
WACC= (wd)(rd)(1 – T) + (wc)(rs)
0.0996 = (0.4)(0.09)(1 – 0.4) + (0.6)rs
0.0996 = 0.0216 + 0.6rs
0.078= 0.6rs
rs= 13%
A series RC circuit contains two resistors and two capacitors. The resistors are 39 Ω and 68 Ω. The capacitors have capacitive reactances of 60 Ω and 50 Ω. The applied voltage is 240 V. What is the voltage drop on the capacitor that has 60 Ω of reactance?
Answer:
Vc = 94 V
Explanation:
In a series circuit, powered by a sinusoidal source, the current is the same across all circuit elements, and can be calculated based on the impedance concept, that takes into account that the voltage and the current in a given circuit element have a phase angle, which means that voltage and current don´t reach to a maximum at the same time.
We define the impedance as the complex sum of the resistive part of the circuit (in which current and voltage be in phase) and the reactive part (in which current and voltage are 90º out of phase), as follows:
Z = R + j (XL - XC)
In order to obtain the magnitude of Z, we can do the following:
Z =√R²+ (Xl-Xc)²
The value of the current can be obtained just dividing the applied voltage between the magnitude of the impedance.
For our circuit, we can get the magnitude of Z as follows:
Z= √(39+68)²+(60+50)² = 154 Ω
⇒ I = V/Z = 240 V / 154 Ω = 1.56 A
In order to get the magnitude of the voltage drop across any circuit element, we just need to apply the generalized version of Ohm´s Law.
For a capacitor, the voltage drop can be calculated as follows:
Vc = I* Xc
⇒ Vc = 1.56 A * 60Ω = 94 V
The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They test several mix designs and chose one that has a sample mean of 900 psi with a standard error of the mean of 10 psi. The team wishes to be very conservative in their design and calculates a one-sided 99% confidence interval on the lower limit. What is this value?
Answer:
874 psi
Explanation:
Given a sample mean (x') = 900,
and a standard error (SE) = 10
At 99% confidence, Z(critical) = 2.58
That gives 99% confidence interval as,
x' ± Z(critical) x SE = 900 ± 2.58 x 10
The value of the lower limit is,
900 - 25.8 = 874.2
≈ 874 psi
A brick of 203 x 102 x 57 mm in dimensions is being burned in a kiln to 1100°C, and then allowed to cool in a room with ambient air temperature of 30°C and convection heat transfer coefficient of 5 W/m2·K. If the brick has properties of rho=1920 kg/m3,Crho= 790 J/Kg·K, and k = 0.90 W/m·K, determine the time required to cool the brick to a temperature difference of 5°C from the ambient air temperature.
Answer:
407 minutes
Explanation:
Step 1: Calculate the volume of the brick
[tex]V = 0.203 X 0.102 X 0.057[/tex]
V = 0.0012 m³
Step 2: Calculate the surface area of the brick
A= 2[(0.203 X 0.102) +(0.203 X 0.057) +(0.102 X0.057)] = 0.08 m²
Step 3: calculate the characteristic length
[tex]L_{C} =\frac{V}{A}[/tex]
[tex]L_{C} = \frac{0.0012}{0.08}[/tex] = 0.015 m
Step 4: calculate the biot number
[tex]B_{i} = \frac{hL_{c} }{k}[/tex]
[tex]B_{i} = \frac{5X0.015 }{0.9}[/tex] = 0.083
⇒Since [tex]B_{i}[/tex] ∠ 0.1, the lumped system analysis is applicable. Then cooling time is determined from
[tex]b = \frac{hA}{\rho c_{p}V } = \frac{h}{\rho c_{p}L_{c} }[/tex]
[tex]b = \frac{h}{\rho c_{p}L_{c} }[/tex]
[tex]b = \frac{5}{1920 X 790 X 0.015}[/tex]
b = 0.0002197 s⁻¹
[tex]\frac{T(t) -T_{o}}{T_{i} - T_{o}} =e^{-bt}[/tex]
[tex]\frac{5}{1100 - 30} =e^{-0.0002197t}[/tex]
Take natural log of both sides
-5.3659 = -0.0002197t
t = 24,424 seconds = 407 minutes
The required time "7 hours".
Air temperature:
Dimension of brick[tex]= 203\times 102\times 57 \ mm\\\\[/tex]
kiln Temperature [tex]\ T_t = 1100^{\circ}\ C \\\\[/tex]
Air temperature of Ambient [tex]\ T_{\infty} = 30^{\circ}\ C\\\\[/tex]
Heat transmission coefficient by convection:
[tex]\to h=5\ \frac{W}{m^2}\ K\\\\[/tex]
Properties of Bricks:
[tex]\to \rho = 1920\ \frac{kg}{m^3}\\\\ \to C_p = 790\ \frac{J}{kg-K}\\\\ \to k=0.9 \frac{W}{m-K}\\\\[/tex]
Calculating the temperature difference:
[tex]\to T_t -T_{\infty} = 5^{\circ}\ C\\\\[/tex]
Where t represents the amount of time needed to cool the brick for a temperature of
[tex]\to T_t = 35^{\circ}\ C\\\\[/tex]
We are familiar with the lumped system analysis for energy balance.
[tex]\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = \frac{hA_s}{\rho V C_p} t \\\\\\[/tex]
Calculating the brick surface area:
[tex]\to A_s = 2(ab + bc +ca)[/tex]
[tex]= 2(0.203\times 0.102 +0.102 \times 0.057 +0.057\times 0.203)\\ \\= 2(0.020706 +0.005814 +0.011571)\\\\ = 2 \times 0.038091 \\\\ =0.076182 \ m^2\\\\[/tex]
Calculating the volume:
[tex]\to V = abc[/tex]
[tex]= 0.057 \times 0.203 \times 0.102 \\\\ = 0.00118\ m^3\\\\[/tex]
We know that:
[tex]\to \ln(\frac{T_t-T_{\infty}}{T_i-T_{\infty}}) = - \frac{hA_s}{\rho V C_p} t \\\\\to \ln(\frac{5}{1100-30}) = \frac{5\times 0.076}{1920 \times 790 \times 0.00118 } t \\\\\to -5.3659=-2.128 \times 10^{-4}\ t\\\\\to t =\frac{-5.3659}{-2.128 \times 10^{-4}}\\\\\to t= 2.521423 \times 10^{4}\ s\\\\\to t=25214.23 \ s\\\\\to t=7.0039527778 \ h\\\\[/tex]
Therefore, the final answer is "7 hours".
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Describe an E2 mechanism
A. Not stereospecific and not concerted
B. Stereospecific and not concerted
C. Stereospecific and concerted
D. Not stereospecific and concerted
Answer:
The answer is C): Stereospecific and concerted.
Explanation:
An E2 mechanism is a single step elimination reaction that is both stereospecific and concerted: it is stereospecific because it can convert the components of the reaction into stereoisomeric products; it is concerted because all bonds are broken and formed during the single step.
A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.
a. What is the molar flow rate of the gas? kmol/h
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C
Answer:
a.6531.53 mole/hr
b. 32.76 degC
c. 3.78 deg.c
Explanation:A gas mixture containing 85.0 mole% N2 and the balance n-hexane flows through a pipe at a rate of 100.0 m3/h. The pressure is 2.00 atm absolute and the temperature is 140.0°C.
a. What is the molar flow rate of the gas? kmol/h
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? °C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75.0% of the hexane? °C
Given V= Volume of gas mixture= 100m3/hr=10^5 Lt/hr P= 2.0 atm and T= 100 deg.c =100+273.15= 373.15K
n= moles of mixture= PV/RT , where R=0.08206 L.atm/mole.K
n= 2.0*10^5/0.08206*373.15)=6531.53 mole/hr
b. To what temperature would the gas have to be cooled at constant pressure in order to begin condensing hexane? A gas mixture containing 85 mole% N2 and the°C
Condensation begins at a point at which the partial pressure of vapor =vapor pressure of liquid at the given temperature
Partial pressure of hexane in the mixture= 0.15*2.0= 0.3 atm
so for saturation to begin, the vapor pressure shoudl correspond to 0.3 atm=30.39 Kpa
Antoine constant for Hexane
lnP (Kpa)= 13.82- 2696/(T-48.833) ( T is in K
ln(30.39 )= 13.8193- 2696/ (T-48.833)
, 3.414= 13,82-2696/(T-48.333)
2696/(T-48.833)= 13.82-3.414=10.405
T-46.833= 2696/10.414=259.08
259.08+46.833 K=305.91k
305.91k-273.15K
32.76C
c. To what temperature would the gas have to be cooled at constant pressure in order to condense 75% of the hexane? °C?
Vapor present in the gas mixture= 25%, Its partial pressure=0.15* 0.25*2.0 =0.075 atm= 7.59Kpa
from ln(7.59)= 13.82- 2696/ (T-48.333)
2.02 = 13.82- 2696/(T-48.333)
2696/(T-48.333)= 11.179
T-48.333= 2696/11.79=228.60
T= 228.6+48.333= 276.93 K= 3.78 deg.c
A 10 m3rigid insulated tank contains air at 1000 kPa, 1000 K. It is connected to an adiabatic reversible turbine that has an exit of 100 kPa.
The air is now allowed to run out through the turbine until a tank pressure of 100 kPais reached and the flow stops.
a) Find the initial and final mass in the tank.
b) How much work did the turbine produce.
Answer:
a) initial mass = 34.85 gm, final mass = 3.48gm
b)
Explanation:
At the initial state, the:
Pressure, p1 = 1000kPa,
Temperature, T1 = 1000K,
Volume, V1 = 10 m³
The mass of air in the tank can be evaluated using;
m1 = (p1 x V1) /(R x T1)
m1 = (1000 x 10)/(287 x 1000)
Therefore, m1 = 34.85g
At the final state,
Pressure, p2 = 100kPa,
Temperature, T2 = 1000K,
Volume, V2 = 10 m³
The mass of air in the tank is given by;
m2 = (p2 x V2) / R x T2
m2 = (100 x 10)/ 287 x 1000
Therefore m2 = 3.48 gm
The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without using trigonometric identities. However, you cannot use the phasor technique in all cases. Select the expressions below for which the phasor technique cannot be used to combine the sinusoids into a single expression.
Check all that apply.
O 45 sin(2500t – 50°) + 20 cos(1500t +20°)
O 25 cos(50t + 160°) + 15 cos(50t +70°)
O 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)
O -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
O 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
The phasor technique for combining sinusoidal functions into a single expression cannot be used when the frequencies within the same expression differ. The expressions 45 sin(2500t – 50°) + 20 cos(1500t +20°) and -100 sin(10,000t +90°) + 40 sin(10,100t – 80°) + 80 cos(10,000t) both contain differing frequencies, and therefore cannot be combined using the phasor technique.
Explanation:In the phasor technique, for the method to be applicable, all sinusoidal functions being combined should have the same frequency. Looking at the expressions given, the phasor technique cannot be applied to these:
45 sin(2500t – 50°) + 20 cos(1500t +20°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)The reasoning for these exceptions is due to discrepancies in the frequencies of the sinusoidal functions within the same expression; in these two cases, the frequencies 2500t and 1500t, as well as 10,000t and 10,100t, do not match, making the phasor technique inapplicable.
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A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. What is the Rankine passive earth pressure on the wall?
The Rankine passive earth pressure on a 12-ft high retaining wall with backfill of granular soil having an internal angle of friction of 30° and a unit weight of 125 pcf is calculated to be 500 psf.
Explanation:The question relates to calculating the Rankine passive earth pressure on a retaining wall that is 12-ft high with backfill of granular soil. The internal angle of friction (φ) provided is 30°, and the unit weight (γ) of soil is 125 pcf (pounds per cubic foot). First, to calculate the passive earth pressure (Ψp), we use the Rankine theory formula: Ψp = γh [1 - sin(φ)]/[1 + sin(φ)], where h is the height of the wall. Substituting the given values, Ψp = 125 * 12 * [1 - sin(30°)]/[1 + sin(30°)]. Since sin(30°) = 0.5, the calculation simplifies to: Ψp = 125 * 12 * [1 - 0.5]/[1 + 0.5], which further simplifies to Ψp = 125 * 12 * 0.5/1.5. Therefore, the Rankine passive earth pressure on the wall amounts to Ψp = 500 psf (pounds per square foot).
Explain why it was important to clean each piece of metal you used with sandpaper prior to doing the experiment. (2) What specific problem might you have encountered had you cleaned the metal surfaces witha soft tissue instead? On the basis of your results from this experiment, what observation(s) did you make that woul allow you to determine the place H2 would take in your relative activity series?
Answer: Metal has a oxide layer when in air. Before an experiment cleaning it with sand paper will remove the leyer of oxide and ensure truer results. Cleaning with tissue paper will not result in removing the layer of oxide.
Shows how active H2 can be.
Explanation:
Following are the responses to the given question:
For question 1:
Sandpaper must be used to polish the metallic surface utilized for the experimentation. Metals oxidize in the environment, forming an oxide coating on the surface thus rendering them unreactive. As a consequence, sand is required to remove this before continuing with the reaction.For question 2:
Then using soft tissue to cleanse a metal's surfaces, the oxides layer on top remains intact. The metal does not fully react underneath the response circumstances. As a consequence, the experimental data were untrustworthy and incorrect.For question 3:
Unless the metal inside the experiment can generate H2, it's much less reactive than Hydrogen. In the relative series, the metal gets put below the hydrogen.When creating the hydrogen, suggesting that metals are acting as a strong reduction agent. Most metals in the range above hydrogen act as oxidizing agents, while most metals below hydrogen act as reductants.Learn more:
brainly.com/question/18364237
1 moles of a gas are initially initially under pressure 4 × 105 Pa at T=300 K. In this problem, you will want to use the following relationships:Definition of Gibbs free energy: G=U−TS+PV Equipartition: U=NDOF2NkT+constEntropy of an ideal gas: S=NklnV+const Ideal gas law: pV=NkTEquation of state for system with excluded volume b: (V−Nb)p=NkT
1)If the gas is ideal, how much does G change when the gas is allowed to expand isothermally until the pressure is reduced to 2 × 105 Pa?
Answer = -1729
2)How much did the chemical potential, ??, change in part (1)? ?final-?initial= ?
3) Now assume the gas has no interaction energy but does have an excluded volume per particle, b=1.9 × 10-28 m3, which means that S = Nk ln(V-Nb) + f(T,N) , where f is some function you won't need to know since T and N don't change. How much does G change when the gas expands isothermally until the pressure is reduced from 4 × 105 Pa to 2 × 105Pa? Gfinal-Ginitial= ?
4) How much did the chemical potential change in the process of part (3)? ?final-?initial= ?
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Refrigerant-134a enters a compressor as a saturated vapor at 160 kPa at a rate of 0.03 m3/s and leaves at 800 kPa. The power input to the compressor is 10 kW. If the surroundings at 20°C experience an entropy increase of 0.008 kW/K, determine (a) the rate of heat loss from the compressor, (b) the exit temperature of the refrigerant, and (c) the rate of entropy generation.
Answer:
a) [tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) T_2 = 36.4 degree C
c) [tex]\Delta S_{gen} = 0.006512 KW/K[/tex]
Explanation:
Given data:
[tex]P_1 = 160 kPa[/tex]
volumetric flow [tex]V_1 = 0.03 m^3/s[/tex]
[tex]P_2 = 800 kPa[/tex]
power input [tex]W_{in} = 10 kW[/tex]
[tex]T_{surr} = 20 degree C[/tex]
entropy = 0.008 kW/K
from refrigerant table for P_1 = 160 kPa and x_1 = 1.0
[tex] v_1 = 0.12355 m^3/kg[/tex]
[tex]h_1 = 241.14 kJ/kg[/tex]
[tex]s_1 = 0.94202 kJ/kg K[/tex]
a) mass flow rate [tex] \dot m = \frac{V_1}{v_1}[/tex]
[tex]\dot m = \frac{0.03}{0.12355} = 0.2428 kg/s[/tex]
heat loss[tex] = T_{surr} \times entropy[/tex]
heat loss[tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) from energy balance equation
[tex]W_{in} 0 \dot Q_{out} = \dot m (h_2 -h_1}[/tex]
[tex]10 - 2.344 = 0.2428 (h_2 - 241.14}[/tex]
[tex]h_2 = 272.67 kJ/kg[/tex]
from refrigerant table, for P_2 = 800 kPa and h_2 = 272.67 kJ/kg
T_2 = 36.4 degree C
c) from refrigerant table P_2 = 800 kPa and h_2 = 272.67 kJ /kg
[tex]s_2 = 0.93589 kJ/kg K[/tex]
rate of entropy
[tex]\Delta S_R = \dot m =(s_2 -s_1)[/tex]
[tex]\Delta S_R = 0.2428 \times (0.93589 -0.94202) = - 0.0014884 kW/K[/tex]
rate of entropy for entire process
[tex]\Delta S_{gen} = \Delta _S_R + \Delta_{surr}[/tex]
[tex]\Delta S_{gen} = 0.0014884 + 0.008 = 0.006512 KW/K[/tex]
A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The allowable stress for the hoops is σallos=11.4 ksi. Determine the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi.
Answers:
31.7 inches
Explanation:
Given:
Diameter = 3ft
Let D = Diameter
So, D = 3ft. (Convert to inches)
D = 3 * 12in = 36 inches
Coss-sectional area of the steel = 0.2in²
Gauge Pressure (P) = 4psi
Stress in Steel (σ)= 11.4ksi
Force in steel = ½ (Pressure * Projected Area)
Area (A) = 2 * Force/Pressure
Also, Area (A) = Spacing (S) * Wood Pipe Diameter
Area = Area
2*Force/Pressure = Spacing * Diameter
Substitute values I to the above expression
2 * Force / 4psi = S * 36 inches
Also
Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel
So, F = 11.4ksi * 0.2in²
F = 11.4 * 10³psi * 0.2in²
F = 2.28 * 10³ psi.in²
So, 2 * Force / 4psi = S * 36 becomes
2 * 2.28 * 10³/4 = S * 36
S = 2 * 2.28 * 10³ / (4 * 36)
S = 4560/144
S = 31.66667 inches
S = 31.7 inches (approximated)
"It is better to be a human being dissatisfied than a pig satisfied; better to be Socrates dissatisfied than a fool satisfied. And if the fool, or the pig, are of a different opinion, it is because they only know their own side of the question. The other party to the comparison knows both sides." This is a very famous quotation from Mill. How would you explain what it says to someone else?
Explanation:
It may be instructive to look at the opposite of the sentence here. Perhaps the smarter creature would be more unhappy when frustrated, recognizing how it gains from happiness relative to the creature with less experience and less knowledge of a situation that does not define it at the moment.
Perhaps the argument is really about the fact that wisdom helps one to hypothetically live in multiple states and a lack of wisdom prevents or fails this possibility.
Your department's laser printer recently began printing a vertical black line near the edge of every printed page. What should you do to resolve the problem?
Answer:
Replace the toner cartridge
Explanation:
solution
when laser printer print black color liner vertical line print it is very frustrating that condition nearly empty toner cartridge
but first we clean the corona wire for that color line
and Reinstall the toner cartridges after Shake the toner cartridge side to side
if problem not solve than Clean the drum unit
and finally if not solve change the toner cartridge
so as that our problem will be resolve
4. Write the command to import the coolfunc() function from the neatomod module in the funpack package and rename it to coolf. Now write the command to import all functions from the neatomod module. What command can we use to obtain a list of all functions inside the module?
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
A rigid tank initially contains 1.4 kg saturated liquid water at 200∘C. At this state, 25 percent of the volume is occupied by water and the rest by air. Now heat is supplied to the water until the tank contains saturatd vapor only. Determine (a) the volume of the tank. (b) the final temperature and pressure. (c) the internal energy change of the water.
Answer:
Volume of tank = 6.4792 × 10 ³ m³
Final temperature = 371°C
Final Pressure = 21.3 kilo pascal
Change in temperature = 1373.54 kilo joule per kilogram.
Explanation:
Saturated liquid water = 1.4 kilogram
T1 = 200°C
At T1, = 25% of liquid = V1 = water
At T1 = 75% = air
T2 = ? , V2 = ? ΔT = ?
Referencing the question,
to be able to calculate the volume of the tank, we need to know the volume of water in the tank.
Therefore, Volume of water = 1.4 kilogram × 0.001157 m³/kg
= 1.6198 × 10³ m ³
Now going back to calculate the volume of the tank,
We have, Volume of tank = 4 × Volume of water
= 4 × 1.6198 × 10³ m ³
6.4792 × 10 ³ m ³
Now, to calculate the pressure and the temperature, we need to calculate the specific volume of the mix,
So we have, α vapor = volume of tank ÷ saturated liquid vapor
we have, 6.4792 × 10 ³ m ³ ÷ 1.4
0.004628 m³ / kilogram
T2 now therefor is 371°C and P2 = 21.3 kilo pascal
Change in temperate of water = T2 - T1
= 2224 - 850.46
1373.54 kilojoule/kilogram
Hydrogen peroxide, H2O2, enters a gas generator at 25 Celsius, 500 kPa, at the rate of 0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa. Determine the power output of the turbine and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2O2 is −187 583 kJ/kmol.
Take a look at the pictures that should help you out.
Ruth owns Skyview, a complex under construction that will include commercial and residential suites, and a parking garage. She allows Town Contracting, the contractor, to complete a stage of the project late. Thiswaives Ruth's right to sue for Select one: a. none of the choices. b. any subsequent breaches. c. any past breaches. d. this delay.
Answer:
Option D
This delay
Explanation:
The contract between Ruth and the contractor is assumed to be oral and since the contractor is supposed to complete a stage of the project when date for completion is over. Ruth can sue the contractor for the delay of the project. The contractor may consequently pay an interest for the delay of the project to Ruth.
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6660 N (1500 lbf). If the length of the rod is 380 mm (15.0 in.), what must be the diameter to allow an elongation of 0.50 mm (0.020 in.)?
Answer:
7.65 mm
Explanation:
Stress, [tex]\sigma=\frac {F}{A}[/tex] where F is the force and A is the area
Also, [tex]\sigma=E\times \frac {\triangle L}{L}[/tex]
Where E is Young’s modulus, L is the length and [tex]\triangle L[/tex] is the elongation
Therefore,
[tex]\frac {F}{A}= E\times \frac {\triangle L}{L}[/tex]
Making A the subject of the formula then
[tex]A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}[/tex]
Since [tex]A=\frac {\pi d^{2}}{4}[/tex] then
[tex]d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm[/tex]