Answer:
commercial freezing
Explanation:
smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected
A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and a quality of 0.2. The tank has a pressure-regulating venting valve that allows pressure to be constant. The tank is subsequently being heated until its content becomes a saturated vapor (of quality 1.0). During heating, the pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. You can neglecting the kinetic and potential energy effects.
Required:
a. Determine the total mass in the tank at the initial and final states, in kg.
b. Calculate the amount of heat (in kJ) transferred from the initial state to the final state.
Answer:
The total mass in the tank = 0.45524 kg
The amount of heat transferred = 3426.33 kJ
Explanation:
Given that:
The volume of the tank V = 0.06 m³
The pressure of the liquid and the vapor of H2O (p) = 15 bar
The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]
By applying the energy rate balance equation;
[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]
where;
[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]
Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]
If we integrate both sides; we have:
[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
We obtain the following data from the saturated water pressure tables, at p = 15 bar.
Since:
[tex]h_e =h_g[/tex]
Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]
[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]
[tex]v_g = 0.1318 \ m^3/kg[/tex]
Hence;
[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]
[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]
[tex]v_1 = 0.02728 \ m^3/kg[/tex]
Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar
[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]
Hence;
[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]
[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]
[tex]u_1 = 1193.428[/tex]
However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:
[tex]m_1 = \dfrac{V}{v_1}[/tex]
[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]
[tex]m_1 = 2.1994 \ kg[/tex]
From the question, given that the final quality; [tex]x_2 = 1[/tex]
[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]
[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]
[tex]v_2 = 0.1318 \ m^3/kg[/tex]
Also;
[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]
[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]
[tex]u_2 = 2594.5 \ kJ/kg[/tex]
Then the final mass can be calculated by using the formula:
[tex]m_2 = \dfrac{V}{v_2}[/tex]
[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]
[tex]m_2 = 0.45524 \ kg[/tex]
Thus; the total mass in the tank = 0.45524 kg
FInally; from the previous equation (1) above:
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]
Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]
Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]
Q = 3426.33 kJ
Thus, the amount of heat transferred = 3426.33 kJ
I need help with simply science
Answer:
mountain ranges may be
Which of the following is an example of a tax
Answer:
A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.
I dont know I asked this to
Explanation:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
/s and at a
velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Figure.
Determine the force acting on the shaft (which is
also the force acting on the bearing of the shaft) in
the axial direction.
Answer:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
Step-by-step solution:
Step 1 of 5
Given data:-
The velocity of water is .
The water flow rate is.
A roadway is to be designed on a level terrain. The roadway id 500 ft. Five cross-sections have been selected at 0 ft, 125 ft, 250 ft, 375 ft, and 500 ft. the cross sections have areas of 130 ft^2, 140 ft^2, 60 ft^2, 110 ft^2, and 120 ft^2. What is the volume needed along this road assuming a 6% shrinkage?
Answer:
51112.5 ft^3
Explanation:
Determine the volume needed along the road when we assume a 6% shrinkage
shrinkage factor = 1 - shrinkage = 1 - 0.06 = 0.94
first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method
Volume between A1 - A2
= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]
= 125 ft * 135 ft^2
= 16875 ft^3
Volume between A2 - A3
= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]
= 125 ft * (200 ft^2 / 2)
= 12500 ft^3
Volume between A3 - A4
= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]
= 125 ft * (170 ft^2 / 2)
= 10625 ft^3
Volume between A4 - A5
(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]
= 125 ft * 115 ft^2
= 14375 ft3
Hence the total volume along the 500 ft road
= ∑ volumes between cross sectional areas
= 16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3
Finally the volume needed along this road is calculated as
Total volume * shrinkage factor
= 54375 * 0.94 = 51112.5 ft^3
The big ben clock tower in london has clocks on all four sides. If each clock has a minute hand that is 11.5 feed in length, how far does the tip of each hand travel in 52 minutes?
Answer:
Updated question
The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?
The distance traveled by the tip of the minute hand of the clock would be 62.59 ft
Explanation:
Let us assume the shape of the clock is circular.
the minute hand is equal to the radius = 11.5 ft
Diameter = radius x 2
Diameter = 11.5 x 2 = 23 ft
The distance traveled by the tip of the minute hand can be calculated thus;
the fraction of the circumference traveled by the minute hand would be;
52/60 = 0.8667
Circumference of the clock would be;
C = pi x d
where C is the circumference
pi is a constant
d is the diameter
C = 3.14 x 23
C = 72.22 ft
Therefore the fraction of the circumference covered by the minute hand would be;
72.22 ft x 0.8667 = 62.59 ft
Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft
Which type of forming operation produces a higher quality surface finish, better mechanical properties, and closer dimensional control of the finished piece?A. Hot working.B. Cold working.
Answer:
Option B (Cold working) would be the correct alternative.
Explanation:
Cold working highlights the importance of reinforcing material without any need for heat through modifying its structure or appearance. Metal becomes considered to have been treated in cold whether it is treated economically underneath the material's transition temperature. The bulk of cold operating operations are carried out at room temperature.The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.
I need help please thank for the help on the last one <3
A cylindrical specimen of Aluminium having a diameter of 12.8 mm and gauge length of 50.8 is pulled in tension. Use the data given below to:A) Plot the data as engineering stress versus engineering strain. B) Compute the modulus of elasticity. C) Determine the yield strength at a strain offset of 0.002. D) Determine the tensile strength of this alloy.E) What is the approximate ductility, in percent elongation?Load (N) Length0 50.8007330 50.85115100 50.90223100 50.95230400 51.00334400 51.05438400 51.30841300 51.81644800 52.83246200 53.84847300 54.86447500 55.88046100 56.89644800 57.65842600 58.42036400 59.182
Answer:
Hello the needed data given is not properly arranged attached below is the properly arranged data
Answer:
b) 62.5 * 10^3 MPa
c) ≈ 285 MPa
d) 370Mpa
e) 16%
Explanation:
Given Data:
cylindrical aluminum diameter = 12.8 mm
Gauge length = 50.8 mm
A) plot of engineering stress vs engineering strain
attached below
B ) calculate Modulus of elasticity
Modulus of elasticity = Δб / Δ ε
= ( 200 - 0 ) / (0.0032 - 0 ) = 62.5 * 10^3 MPa
C) Determine the yield strength
at strain offset = 0.002
hence yield strength ≈ 285 MPa
D) Determine tensile strength of the alloy
The tensile strength can be approximated at 370Mpa because that is where it corresponds to the maximum stress on the stress vs strain ( complete plot )
E) Determine approximate ductility in percent elongation
ductility in percent elongation = plastic strain at fracture * 100
total strain = 0.165 , plastic strain = 0.16
therefore Ductility in percent elongation = 0.16 * 100 = 16%
What is difference between a backdoor, a bot, a keylogger, and psyware,a nd a rootkit? Can they all present in the same malware?
Answer:
Yes, they can all be present in the same malware because each of them perform slightly differing functions.
Explanation:
Backdoor is a software which when placed into your computer will permit hackers to easily gain reentry into your computer. This can happen even after you have already patched the flaw that they have used to hack your system before.
A bot is a program that does the same task in a continuous manner akin to when you use a blender by pressing the button to blend what you have put into it.
A keylogger is a part of a hidden software that monitors and records everything you type on your computer keyboard after which it writes it onto a file, with the hopes of capturing relevant information such as your bank account number and even passwords and other sensitive means of identification.
A Spyware is somehow similar to a keylogger just that it steals information from your computer and sends it to someone else.
A root kit is a bad software that is capable of modifying the operating system or other privileged access devices in order to gain continuous access into your system for the purpose of gathering of information and/or reducing the system’s functionality.
Yes, they can all be present in the same malware because each of them perform slightly differing functions.
Consider a pan of water being heated (a) by placing it on an electric range and (b) by placing a heating element in the water. Which method is a more efficient way of heating water? Explain.
Answer:
Method B is the more efficient way of heating the water.
Explanation:
Method B is more efficient because by placing a heating element in the water as in described in method B, the heat that is lost to the surroundings is minimized which implies that more heat is supplied directly to the water. Therefore, more heating is achieved with a lesser amount of electrical energy input. Whereas placing the pan on a range means more heat losses to the surrounding and as such it will take a longer time for the water to heat up and also take more electrical energy.
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
dutile is the correct answer
A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
a. 120
b. 55
c. 175
d. 230
An equal-tangent sag vertical curve is designed with a PVC as station 109+00 and elevation 950ft, the PVI has a station of 110+77 and elevation of 947.34ft, and the low point at station 110+50. Determine the design speed of the cure.
Answer:
K = 96 and the design speed of the curve = 50mph
Explanation:
109+00 = 10900
Elevation = 950ft
110+77 = 11077
Elevation = 947.34ft
Station of low point = 110+50 = 11050
To get grade of curve
Gi = 947.34-950/11077-10900
= -2.66/177
= -0.015x100
= -1.5%
Locate of low point (XL)
= 11050-10900
= 150
To get the value of K
XL = |GL| x K
When we substitute values
150 = 1.5 x K
150 = 1.5K
K = 150/1.5
K = 100
The suitable and most nearest value is K = 96
Then we use the standard chart to get the design speed for K = 96
On this chart, the design speed for the curve = 50mph
Therefore K = 96 and speed = 50mph
If you make a mistake in polarity when measuring the value of DC voltage in a circuit with a digital VOM, what will happen? A. The meter will be damaged. B. The meter will read positive voltage only. C. The meter will display a negative sign. D. The meter will display OL which states an overload condition.
Answer:
C. The meter will display a negative sign.
Explanation:
If you use an analog voltmeter and you measure voltage with reverse polarity you will damage it. But in this case we are using a digital multimeter. This kind of multimeter is designed to be able to deal with positive and negative voltages
What is the Bernoulli formula?
Answer:
P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
Sieve size Weight retained (g) No. 4 59.5 No. 8 86.5 No. 16 138.0 No. 30 127.8 No. 50 97.0 No. 100 66.8 Pan 6.3
Solution :
Sieve Size (in) Weight retain(g)
3 1.62
2 2.17
[tex]$1\frac{1}{2}$[/tex] 3.62
[tex]$\frac{3}{4}$[/tex] 2.27
[tex]$\frac{3}{8}$[/tex] 1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan 6.3 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]
[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]
x = 0.2634 mm
Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]
x = 1.6317 mm
[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]
Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]
[tex]$Cu= \frac{1.6317}{0.2643}$[/tex]
Cu = 6.17
Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]
[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm
Refrigerant-22 absorbs heat from a cooled space at 50°F as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at 10°F at a rate of 0.08 lbm/s with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine:
a. The rate of cooling provided, in Btu/h.
b. The rate of exergy destruction in the evaporator.
c. The second-law efficiency of the evaporator.
Take T0 = 77°F. The properties of R-22 at the inlet and exit of the evaporator are: h1 = 107.5 Btu/lbm, s1 = 0.2851 Btu/lbm·R, h2 = 172.1 Btu/ lbm, s^2 = 0.4225 Btu/lbm·R.
Answer:
a) the rate of cooling provided is 18604.8 Btu/h
b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) the second-law efficiency of the evaporator is 37.39%
Explanation:
Given that;
Temperature of sink TL = 50°F = 510 R
Temperature at evaporator inlet TI = 10°F = 470 R
mass flow rate m" = 0.08 lbm/s
quality of refrigerant at evaporator inlet x1 = 0.3
quality of refrigerant at evaporator exit x2 = 1.0
T₀ = 77°F = 537 R
h1 = 107.5 Btu/lbm
s1 = 0.2851 Btu/lbm·R,
h2 = 172.1 Btu/ lbm,
s2 = 0.4225 Btu/lbm·R.
a) rate of cooling provided, in Btu/h.
QL = m"( h2 - h1)
we substitute
QL = 0.08( 172.1 - 107.5
= 0.08 × 64.6
= 5.168 Btu/s
we convert to Btu/h
5.168 × 60 × 60
QL = 18604.8 Btu/h
Therefore the rate of cooling provided is 18604.8 Btu/h
b) The rate of exergy destruction in the evaporator
Entropy generation can be expressed as;
S_gen = m"(s2 - s1) - QL/TL
so we substitute
S_gen = 0.08( 0.4225 - 0.2851 ) - 5.168 / 510
= 0.010992 - 0.01013
S_gen = 0.00086 Btu/ibm.R
now the energy destroyed expressed as;
X_dest = T₀ × S_gen
so
X_dest = 537 × 0.00086
X_dest = 0.46 Btu/Ibm
Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) The second-law efficiency of the evaporator.
Energy expended is expressed as;
X_exp = m"(h1 - h2) - m"T₀(s1 - s2)
we substitute
= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )
= -5.168 - [ - 5.9027)
= -5.168 + 5.9027
= 0.7347 Btu/s
Now second law efficiency is expressed as;
nH = 1 - (X_dest / X_esp)
= 1 - ( 0.46 / 0.7347 )
= 1 - 0.6261
= 0.3739
nH = 37.39%
Therefore the second-law efficiency of the evaporator is 37.39%
please help i have no xlue
The structure of a house is such that it loses heat at a rate of 3800 kJ/h per C di erence between the indoors and outdoors. A heat pump that requires a power input of 4 kW is used to maintain this house at 24C. Determine the lowest outdoor temperature for which the heat pump can meet the heating requirements of this house.
Answer:
-9.5° C
Explanation:
See attachment for calculations.
On the concluding parts, from the attachment, we have that
√[(297000 * 4)/(1056)] = 297 - T(l), and solving further, we get
297 - T(l) = √(1188000/1056)
297 - T(l) = √1125
297 - T(l) = 33.5
T(l) = 297 - 33.5
T(l) = 263.5
When you convert back to °C, we have
263.5 - 273 = -9.5° C
An unknown impedance Z is connected across a 380 V, 60 Hz source. This causes a current of 5A to flow and 1500 W is consumed. Determine the following: a. Real Power (kW) b. Reactive Power (kvar) c. Apparent Power (kVA) d. Power Factor e. The impedance Z in polar and rectangular form
Answer:
a) Real Power (kW) = 1.5 kW
b) Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA) is 1.9 KVA
d) the Power Factor cos∅ is 0.7894
e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
Explanation:
Given that;
V = 380v
i = 5A
P = 1500 W
determine;
a) Real Power (kW)
P = 1500W = 1.5 kW
therefore Real Power (kW) = 1.5 kW
b) Reactive Power (kvar)
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
∅ = cos⁻¹ (0.7894)
∅ = 37.87°
Q = VIsin∅
Q = 380 × 5 × sin( 37.87° )
Q = 1.1663 KVAR
Therefore Reactive Power (kvar) is 1.1663 KVAR
c) Apparent Power (kVA)
S = P + jQ
= ( 1500 + J 1166.3 ) VA
S = 1900 ∠ 37.87° VA
S = 1.9 KVA
Therefore Apparent Power (kVA) is 1.9 KVA
d) Power Factor
p = V×i×cos∅
cos∅ = p / Vi
cos∅ = 1500 / ( 380 × 5 ) = 0.7894
Therefore the Power Factor cos∅ is 0.7894
e) The impedance Z in polar and rectangular form
Z = 380 / ( S∠-37.87) = V/I
Z = ( 60 + j 46.647) Ω
Z = 76 ∠ 37.87° Ω
Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω
In beams, why is the strain energy from bending moments much bigger than the strain energy from transverse shear forces? Choose one or more of the following options.
a) The stresses due to bending moments is much more than the stresses from transverse shear.
b) The strains due to bending moments is much more than the strains from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Answer:
a) The stresses due to bending moments is much more than the stresses from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Explanation:
Strain in an object suspended is a function of the stress which the suspended body passed through. The stress which is the function of the force experienced by the body over a given area helps is straining the moment. This lead to the strain energy from bending moment being greater than the strain energy from a transverse shear force.
Which of the following is not one of the common classifications of product liability defects? A. Manufacture B. Materials C. Packaging D. Both "Materials" and "Packaging" E. Design
Answer:
D. Both "Materials" and "Packaging"
Explanation:
Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.
There are majorly three product defects. They are :
1. Manufacturing defect
2. Design defect
3. Marketing defect
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?
Answer:
[tex]10.8\ \text{lb/ft^2}[/tex]
[tex]101.96\ \text{lb/ft}^2[/tex]
Explanation:
[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]
[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]
[tex]v_2=0[/tex]
[tex]P_1=0[/tex]
[tex]h_1=h_2[/tex]
From Bernoulli's law we have
[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]
The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]
Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]
[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]
The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]
A spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
Answer:
0.75 kg
Explanation:
c = Damping coefficient = 3 kg/s
x = Displacement of spring = 0.5 m
F = Force = 1.5 N
From Hooke's law we get
[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]
In the case of critical damping we have the relation
[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]
Mass that would produce critical damping is 0.75 kg.
0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.
What is zero velocity?A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.
For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.
Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.
Thus, it is 0.75 kg.
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7. The surface finish for the cylinder walls usually depends on the
O A. type of engine oil used.
O B. sharpness of the cylinder bore edges.
O C.type of piston rings used
O D. cylinder wall-to-piston clearance.
The structure of PF3(C6H5)2 is trigonal bipyramidal, with one equatorial and two axial F atoms which interchange positions when heated. Describe the low- and high- temperature 31P and 19F NMR spectra.
Answer:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Explanation:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Tubular centrifuge is used for recovering cells 60% of the cells or recover data flow rate of 12 l/min with a rotational speed of 4000 RPM what is the RPM to increase the recovery rate of the cells to 95% at the same flow rate
Answer:
The RPM to increase the recovery rate of the cells to 95% at the same flow rate is 6,333.3 RPM.
Explanation:
If the tubular centrifuge rotates at about 4,000 revolutions per minute to recover 60% of the cells, in case of wanting to recover 95% of the cells, the following calculation must be carried out to determine the required number of revolutions per minute:
60 = 4,000
95 = X
((95 x 4,000) / 60)) = X
(380,000 / 60) = X
6,333.3 = X
Therefore, as the calculation emerges, the tubular centrifuge will need to rotate at about 6,333.3 revolutions per minute to recover 95% of the cells in the same time.
What is computer programming
Answer:
Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.
Explanation:
Hope dis helps! :)
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Red 5 1 or (b) transition to turbulence, Red 5 250,000
Answer:
a. 4[tex]\mu m[/tex]
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4[tex]\mu m[/tex]
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m