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
Consider a 3-phase, 4 Pole, AC induction motor being driven at 75 Hz. Initially, the motor is connected to a mechanical load that requires 1,816 W of power. While connected to this load, the motor spins at 1,996 rpm. Now a new load is connected that requires 2,334 W of power. What is the new slip on the motor for the new load assuming that the motor is operating in its normal operating range?
Answer:
New slip = 0.145
Explanation:
Revolutions per minute rpm = 1996
Synchronous speed Ns= 120 * f / p
= 120 * 75 / 4 = 2250
Hence initial slip S1 = 2250 - 1996 / 2250 = 0.113
From output power = [tex] 3*S*E^2*R2 [/tex]
Hence at constant E and R, output power depends on slip
So for new power and new slip to new initial power and initial slip,
Hence 2334/1816 = S2/ S1
New slip S2 = 2334 * 0.113 / 1816 = 0.145
Hence, new speed = Ns ( 1 - S2 )
= 2250*(1 - 0.145)
= 1924 rpm
Marco is a franchisee with Daggies, a chain of sandwich shops. His business was doing well until several Daggies franchisees got in trouble and were forced to close their shops. Soon afterward, Marco's business deteriorated and he too was forced to close. This is an example of:
Answer:
The coattail effect
Explanation:
Clearly, the Daggies franchise were a huge part of Marco's success and attracted many customers for the business. This is why Marco deteriorated as soon as many Daggies franchisees closed.
The coattail effect is the phenomenon where an influencing member in a party (franchisee in this case) contributes largely to the success of another, which is the case with Marco and Daggies
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
For a two-source network, if current produced by one source is in one direction, while the current produced by the other source is in the opposite direction through the same resistor, the resulting current is? a. The product of the two and the direction of the smaller b. The difference between the two and has the same direction. c. The sum of the two and the direction of either d. The average of the two and the direction of the largest
Answer:
c. The sum of the two and the direction of either.
Explanation:
If both sources are linear, we can apply the superposition theorem, which in this case, states simply that the resulting current is just the algebraic sum of both currents.
This is due to any of them, is independent from the other, so we can calculate her influence without any other source present.
Assuming that both sources are ideal (no shunt resistances) , we can apply superposition, removing all sources but one each time, just replacing the sources by an open circuit.
Let's suppose that we have a 5A source flowing to the rigth, and a 3A source in the opposite direction.
Applying superposition, we have:
I₁ = 5A (the another source is removed)
I₂ = -3A (The minus sign indicates that is flowing in the opposite direction, 5A source is removed)
I = I₁ + I₂ = 5A + (-3A) = 2 A
As it can be seen , the resulting current is the sum (algebraic) of both current sources, and has the direction of either of the sources, depending on the absolute value of the current sources.
(If the directions of I₁ and I₂ were inverted, with the same absolute values, the direction would be the opposite).
We could have arrived to the same result applying KCL.
The simple majority decision rule may generate results that are
Answer:
Efficient when the marginal benefits of project = marginal costs of project.
Explanation:
Majority Decision Rule:
Majority decision rule is based on the notion of equality. An alternative is selected which has majority of votes. The simple majority decision rule may generate efficient results if the marginal benefits of a project are equal or greater than the marginal costs of the project.
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.
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]
What is the correct statement regarding the stress over the section of a shaft in torsion?
a) A normal stress that varies linearly over the section
b) A normal stress that is uniform over the section.
c) A shear stress that is uniform over the section.
d) A shear stress that varies linearly over the section
Answer:
d) A shear stress that varies linearly over the section.
Explanation:
Given that shaft is under pure torsion
As we know that relationship between shear stress in the shaft and radius given as
[tex]\dfrac{T}{J}=\dfrac{\tau}{r}[/tex]
For solid shaft
[tex]J=\dfrac{\pi}{64}d^4[/tex]
Therefore shear stress given as
[tex]\tau=\dfrac{T}{J}\times r[/tex]
T=Applied torque on the shaft
τ=Shear stress at any radius r
From the above equation we can say that shear stress is vaying linearly with the radius of the shaft
Therefore the answer will be d.
Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity is 528 m/s. Determine the static pressure and temperature of the air at this state. The properties of air at an anticipated average temperature of 600 K are cp = 1.051 kJ/kg·K and k = 1.376.
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
[tex]T_0 = T+\frac{V^2}{2c_p}[/tex]
Where
T = Static temperature
V = Velocity of Fluid
[tex]c_p =[/tex] Specific Heat
Re-arrange to find the static temperature we have that
[tex]T = T_0 - \frac{V^2}{2c_p}[/tex]
[tex]T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})[/tex]
[tex]T = 672.88K[/tex]
Now the pressure of helium by using the Adiabatic pressure temperature is
[tex]P = P_0 (\frac{T}{T_0})^{k/(k-1)}[/tex]
Where,
[tex]P_0[/tex]= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
[tex]P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}[/tex]
[tex]P = 0.399Mpa[/tex]
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.
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 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).
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".
Find out more information about air temperature here:
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Discuss how, as a safety professional, you would respond to the overlap in Occupational Safety and Health Administration (OSHA) standards and state or local building, electrical, and life safety codes. What would be the most significant challenges?
Answer:
Explanation:
As a security professional, I will respond positively to the OSHA requirements overlap. OSHA guidelines are meant to provide general guidance to all members of various entities throughout the country, while local or state codes also ensure compliance with laws unique to their areas, taking into account workplace safety and security.
OSHA accepts the security codes of the state building To the degree that such codes comply with OSHA regulations, such as BOCA. All the codes and regulations for local, state-owned construction, electrical and life protection are under the same umbrella. Generally, all security protocols and specifications are in accordance with OSHA guidelines. Nonetheless, certain points will overlap, while localized codes will also be addressed to a particular community or state that may
Employers are not required to keep a record of an employee who has the flu.
a. True
b. False
Employers aren't required to keep records specifically for an employee with the flu, unless under certain conditions related to workplace illnesses. Reasonable exceptions to the FOIA include the protection of sensitive personal information such as government employees' medical records.
Explanation:Employers are not generally required to keep records of an employee who has simply contracted the flu. However, certain regulations may apply if the illness could be work-related or if it pertains to a larger public health concern that requires tracking. In the context of the Freedom of Information Act (FOIA), the question of keeping medical records would fall under exemptions related to personal privacy. For instance, medical records for government employees would be a reasonable exception to FOIA, as they contain sensitive personal information that is protected from public disclosure.
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 minimum requirements for engineering documents are enumerated in
a. The Florida Building Code.
b. Chapter 471, F.S.
c. Engineer's Responsibwty Role
d. Role 61615-21, F.A.C, relating to 'Seals.
Answer:
The answer will be Rule 61G15-23 F.A.C, relating to Seals.
Explanation:
According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23' from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.
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 heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequality of Clausius. Which process is possible. What is the maximum power available?
a) Pout = 3 kW (power out)
b) Pout = 2 kW
c) Carnot Cycle.
Answer:
Explanation:
Given
[tex]T_h=250^{\circ}C\approx 523\ K[/tex]
[tex]T_L=30^{\circ}C\approx 303\ K[/tex]
[tex]Q_1=6 kW[/tex]
From Clausius inequality
[tex]\oint \frac{dQ}{T}=0[/tex] =Reversible cycle
[tex]\oint \frac{dQ}{T}<0[/tex] =Irreversible cycle
[tex]\oint \frac{dQ}{T}>0[/tex] =Impossible
(a)For [tex]P_{out}=3 kW[/tex]
Rejected heat [tex]Q_2=6-3=3\ kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K[/tex]
thus it is Impossible cycle
(b)[tex]P_{out}=2 kW[/tex]
[tex]Q_2=6-2=4 kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K[/tex]
Possible
(c)Carnot cycle
[tex]\frac{Q_2}{Q_1}=\frac{T_1}{T_2}[/tex]
[tex]Q_2=3.47\ kW[/tex]
[tex]\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}[/tex]
[tex]=\frac{6}{523}-\frac{3.47}{303}=0[/tex]
and maximum Work is obtained for reversible cycle when operate between same temperature limits
[tex]P_{out}=Q_1-Q_2=6-3.47=2.53\ kW[/tex]
Thus it is possible
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.
An air-conditioning system operates at a total pressure of 95kPa and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100°C. Air enters the heating section at 10°C and 70% relative humidity at a rate of 35 m^3/min, and it leaves the humidifying section at 20°C and 60 percent relative humidity. Determine:
(a) the temperature and relative humidity of air when it leaves the heating section.
(b) the rate of heat transfer in the heating section.
(c) the rate at which water is added to the air in the humidifying section.
Answer:
The temperature and relative humidity when it leaves the heating section = T2 = 19° C and ∅2 = 38%
Heat transfer to the air in the heating section = Qin = 420 KJ/min
Amount of water added = 0.15 KG/min
Explanation:
The Property of air can be calculated at different states from the psychometric chart.
At T1 = 10° C and ∅ = 70%
h1 = 87 KJ/KG of dry air
w1 = 0.0053 kg of moist air/ kg of dry air
v1 = 0.81 m^3/kg
AT T3 = 20° C , 3 ∅ = 60%
h3 = 98 KJ/KG of dry air
w3 = 0.0087 kg of moist air/ kg of dry air
The moisture in the heating system remains the same when flowing through the heating section hence, (w1 = w2)
The mass flow rate of dry air,
m1 = V'1/V1 = 35/0.81
m1 = 43.21 kg/min
By balancing the energy in heating section we get:
mwhw + ma2h2 = mah3
(w3 -w2)hw + h2 = h3
h2 = h3 - (w3 -w2)hw @ 100 C
Hence, hw = hg @ 100 C and w2 = w1
h2 = h3 - (w3 -w2) hg @ 100 C
h2 = 98 - ( 0.0087 - 0.0053) * 2676
h2 = 33.2 KJ/KG
The exit temperature and humidity will be,
T2 = 19.5° C and 2 ∅ = 37.8%
(b) Calculating the transfer of heat in the heating section
Qin = ma(h2 -h1) = 43.21(33.2 - 23.5)
Qin = 420 KJ/min
(c) Rate at which water is added to the air in the humidifying section,
mw = ma(w3 - w2) = (43.2)(0.0087 - 0.0053)
mw = 0.15 KG/min
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.
When implementing a safety and health program, management leadership does not need employee participation a True b) False Effective management of worker safety and health programs has O a) Stopped all on-site injuries o b) Improve employee productivity and morale in the workplace c)Cost companies more money O d)Become a waste of time Nearly 3 of all serious occupational injuries and illnesses stem from overexertion of repetitive motion e a1/2 O b) 1/4 c)All o d) 1/3 Training is a way for employers to provide 4 to enable employees to protect themselves and others from injuries O a) Ideas b) Tools o cInteractions O d) Money Under OSHA, employees are not protected from discrimination when reporting a work-related injury, illness, or fatality 5 a True b) False
Answer: I have answered the questions in rephrased sentences as below;
When implementing a safety and health program, management leadership need employee participation. Effective management of worker safety and health programs has improved employee productivity and morale in the workplace.
Nearly a third of all serious occupational injuries and illnesses stem from overexertion of repetitive motion.
Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.
Under OSHA, employees are protected from discrimination when reporting a work-related injury, illness, or fatality.
Explanation: All personnel including management & employees must be directly involved when workplace HSE policies are being made & reviewed. This is because everyone in the work environment is impacted one way or the other when incidents occur.
Training & Reporting are key responsibilities of managers, employers & supervisors, so it is mandatory to be done without discrimination so as to foster employees happiness which ultimately lead to zero incidents & increased productivity & profit.
"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.
A mitochondrial membrane complex consisting of ATP synthase, adenine nucleotide translocase (ATP–ADP translocase), and phosphate translocase functions in oxidative phosphorylation. Adenine nucleotide translocase, an antiporter located in the inner mitochondrial membrane, moves ADP into and ATP out of the matrix. Phosphate translocase is also located in the inner mitochondrial membrane. It transports H + ions and phosphate H 2 PO − 4 ions into the matrix. The energy derived from the movement of H + ions down an electrochemical gradient from the intermembrane space into the matrix is used to drive the synthesis of ATP. How many H + ions must be moved into the matrix for the synthesis of 1 ATP? number of H + ions:'
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 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]
True/False
If a contractor chooses to use their own quality system, they must demonstrate compliance with the established military standard through formal third party certification. [Identify DoD policy regarding Basic Quality Systems and the role of ISO 9001.
Answer:
false jdbebheuwowjwjsisidhhdd
A steam turbine in a power plant receives 5 kg/s steam at 3000 kPa, 500°C. Twenty percent of the flow is extracted at 1000 kPa to a feed water heater, and the remainder flows out at 200 kPa. Find the two exit temperatures and the turbine power output.
Answer:
The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW
Explanation:
If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.
In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg
For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg
For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg
For the power output, we have to multiply the steam flow with the enthalpic jump.
The addition of the 2 jumps is the total power output.
Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types.
Drexel University Financial Office has made contracts with several local banks to help students to easily obtain small loans up to $10,000 for qualified students for each year. A student can apply for loans up to 3 banks each year. For each application, student must indicate bank name, loan amount, and requested date. Obviously, a bank approves some student loans and also decline some loans depending on the student status (which is beyond the scope of this database). For each approved loan, there is a loan#, interest rate, approved amount, monthly payment amount, and the beginning date of loan payments. For each loan, the office also keeps tracks of history of payments to the loan, including payment date and payment amount. The office will record student ID, name, major, and year. For banks, we just keep bank number and name.
You may add other unstated, but common-sense oriented, facts to the ERD, but you must represent the facts stated above.
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.
Make two lists of applications of matrices, one for those that require jagged matrices and one for those that require rectangular matrices. Now argue whether just jagged, just rectangular, or both should be included in a programming language.
Matrices, particularly jagged and rectangular ones, serve different functions in programming. Jagged matrices handle non-uniform data and graphs, while rectangular matrices are necessary for graphics and linear algebra. Both should be included in programming languages to provide flexibility and cater to various applications.
Explanation:Applications of Matrices in ProgrammingMatrices are a fundamental aspect of both mathematics and computer science. They are primarily used for representing and manipulating linear transformations, data, and relations. Matrices come in various forms, with two common types being jagged matrices and rectangular matrices.
When considering whether to include jagged or rectangular matrices, or both, in a programming language, it's essential to recognize the different use cases each type serves. Inclusion of both types allows for greater flexibility and functionality. A programming language that supports both can cater to a wider range of applications and user needs. Nullable matrices can be used when necessary to mimic jaggedness within rectangular matrix structures, offering a middle ground.
Including both jagged and rectangular matrices in a programming language offers flexibility for diverse applications, optimizing memory usage and performance across various data structures and operations.
Applications of Matrices
Applications that Require Jagged Matrices
1. Sparse Data Representation: Efficient storage of data where the majority of elements are zero, such as social network adjacency matrices.
2.Graph Representation: Storing graphs where the number of edges varies widely between nodes.
3. Image Processing: Non-uniform image sampling or varying resolution images.
4. Variable-Length Data: Managing records of varying lengths in databases or documents.
5. Numerical Solutions: Adaptive mesh refinement in computational fluid dynamics.
Applications that Require Rectangular Matrices
1. Linear Algebra: Solving systems of linear equations, eigenvalue problems.
2. Data Analysis: Handling datasets in machine learning, where data points have the same number of features.
3. Computer Graphics: Transformations and projections in 2D and 3D graphics.
4. Control Systems: State-space representations of dynamic systems.
5. Signal Processing: Filtering, convolution operations in audio and image processing.
Arguments for Including Matrix Types in a Programming Language
Just Jagged Matrices
- Pros:
- Flexibility in handling datasets with variable row lengths.
- Efficient memory usage for sparse or highly irregular data.
- Cons:
- Increased complexity in implementation and usage, as many linear algebra operations assume rectangular structure.
- Can be less performant for operations that benefit from contiguous memory storage.
Just Rectangular Matrices
- Pros:
- Simplicity in usage and implementation for most standard applications.
- Better support for linear algebra operations and optimizations in libraries.
- Cons:
- Inefficient memory usage for sparse or variable-length data, leading to wasted space.
- Lack of flexibility for applications requiring variable row lengths.
Both Jagged and Rectangular Matrices
- Pros:
- Flexibility to choose the most appropriate type based on the specific application requirements.
- Optimal memory usage and performance by using jagged matrices for sparse data and rectangular matrices for linear algebra operations.
- Cons:
- Increased complexity in the programming language and potential confusion for users in choosing the correct type.
- Possible performance overhead in managing two types of matrices.
Including both jagged and rectangular matrices in a programming language provides the greatest flexibility and efficiency across a wide range of applications. While it adds some complexity, the benefits of being able to choose the optimal structure for specific tasks outweigh the drawbacks. This approach ensures that the language can handle diverse data structures and operations effectively, catering to both irregular and uniform datasets.
A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. Find the power (in Watts [W]) the runner is exerting while running. b) Find the total energy (in Joules [J]) exerted by the runner in a 15 km run.c) How many Milky Way (Original Single 52.2g) chocolate bars does the runner need to buy to supply the amount of energy to complete a half-marathon (13.1 miles)?
Answer:
a) power = 929.78W
b) energy = 3,347,200J
c) 5 milkyway bars
Explanation:
a) Power is the rate of use of energy per unit time. This is given in the question as 800 kilocalories per hour. Part a) requires the power in Watts (W) which is equivalent to Joules per second. Thus, kilocalories per hour need to be converted to Joules per second:
[tex]1kcal = 4184J[/tex]
[tex]1hour = 3600s[/tex]
[tex]power=\frac{energy}{time}[/tex]
[tex]power=\frac{800kcal*\frac{4184J}{kcal}}{1hour*\frac{3600s}{hour}}[/tex]
[tex]power=\frac{3347200}{3600}[/tex]
[tex]power=929.78J/s=929.78W[/tex]
b) Total time required for a 15km run can be calculated by the speed of the runner (15km/h):
[tex]time=\frac{distance}{speed}[/tex]
[tex]time=\frac{15km}{15km/h}[/tex]
[tex]time=1hour=3600s[/tex]
The energy exerted over this time can be found by:
[tex]energy=power*time[/tex]
[tex]energy=929.78J/s*3600s[/tex]
[tex]energy=3,347,200J[/tex]
c) Total time required for a 13.1mile run can be calculated by the speed of the runner (15km/h):
[tex]1mile=1.6km[/tex]
[tex]time=\frac{distance}{speed}[/tex]
[tex]time=\frac{13.1mile*\frac{1.6km}{mile}}{15km/h}[/tex]
[tex]time=\frac{20.96}{15}[/tex]
[tex]time=1.40h=5030.4s[/tex]
The energy exerted over this time can be found by:
[tex]energy=power*time[/tex]
[tex]energy=929.78J/s*5030.4s[/tex]
[tex]energy=4,677,165J[/tex]
Assuming one Milky Way (Original Single 52.2g) has 240 kilocalories
[tex]number=\frac{energy}{energy_{milkyway}}[/tex]
[tex]number=\frac{4677165}{240kcal*\frac{4184J}{kcal}}[/tex]
[tex]number=\frac{4677165}{1004160}[/tex]
[tex]number=4.65[/tex]
5 milkyway bars are needed