Answer:
D = 1311.94 lb/ft
Explanation:
We are given the velocity as;
V = 55 mph.
First of all, let's convert it to ft/s
V = 55 × (5280/3600) ft/s
V = 80.67 ft/s
The equation for the aerodynamic drag force on the car is given as;
D = C_d•½ρ•V²•A
Where;
C_d is drag coefficient = 0.21
ρ is density of air
V is velocity = 80.67 ft/s
A is area 55 ft²
Now, from tables, the density of air under S.T.P condition is 1.225 kg/m³. Converting to lb/ft³ gives; 0.0624 lb/ft³
Plugging in the relevant values, we have;
D = 0.21•½•0.0624•80.67²•30
D = 1311.94 lb/ft
An undersea research chamber is spherical with an external diameter of 3.50 mm . The mass of the chamber, when occupied, is 21700 kg. It is anchored to the sea bottom by a cable. Find the followings
Required:
a. The buoyant force on the chamber.
b. The tension in the cable?
Answer:
a. The buoyant force on the chamber is 220029.6 N
b. The tension in the cable is 7369.6 N
Explanation:
The diameter of the sphere cannot be in millimeter (mm), if the chamber must occupy a big mass as 21700kg
Given;
diameter of the sphere, d = 3.50 m
radius of the sphere, r = 1.75 mm = 1.75 m
mass of the chamber, m = 21700 kg
density of water, ρ = 1000 kg/m³
(a)
Buoyant force is the weight of water displaced, which is calculated as;
Fb = ρvg
where;
v is the volume of sphere, calculated as;
[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \pi (1.75)^3\\\\V = 22.452 \ m^3[/tex]
Fb = 1000 x 22.452 x 9.8
Fb = 220029.6 N
(b)
The tension in the cable will be calculated as;
T = Fb - mg
T = 220029.6 N - (21700 x 9.8)
T = 220029.6 N - 212660 N
T = 7369.6 N
The velocity field of a flow is given by V = 2x2 ti +[4y(t - 1) + 2x2 t]j m/s, where x and y are in meters and t is in seconds. For fluid particles on the x-axis, determine the speed and direction of flow
Answer:
Explanation:
The value of a will be zero as it is provided that the particle is on the x-axis.
Calculate the velocity of particles along x-axis.
[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]
Substitute 0 for y.
[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]
Here,
[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]
Calculate the magnitude of vector V .
[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]
Substitute
[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]
[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]
The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]
Calculate the direction of flow.
[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]
Here, θ is the flow from positive x-axis in a counterclockwise direction.
Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.
[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]
The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.
A certain heat pump produces 200 kW of heating for a 293 K heated zone while only using 75 kW of power and a heat source at 273 K. Calculate the COP of this device as well as the theoretical maximum COP
Answer:
COP(heat pump) = 2.66
COP(Theoretical maximum) = 14.65
Explanation:
Given:
Q(h) = 200 KW
W = 75 KW
Temperature (T1) = 293 K
Temperature (T2) = 273 K
Find:
COP(heat pump)
COP(Theoretical maximum)
Computation:
COP(heat pump) = Q(h) / W
COP(heat pump) = 200 / 75
COP(heat pump) = 2.66
COP(Theoretical maximum) = T1 / (T1 - T2)
COP(Theoretical maximum) = 293 / (293 - 273)
COP(Theoretical maximum) = 293 / 20
COP(Theoretical maximum) = 14.65
Decompose the signal (1+0.1 cos5t) cos100t into a linear combination of sinusoidal functions, and find the amplitude, frequency, and phase of each component. Hint: use the identity for cosacosb.
Answer:
amplitudes : 1 , 0.05, 0.05
frequencies : 50/[tex]\pi[/tex], 105/[tex]2\pi[/tex], 95/2[tex]\pi[/tex]
phases : [tex]\pi /2 , \pi /2 , \pi /2[/tex]
Explanation:
signal s(t) = ( 1 + 0.1 cos 5t )cos 100t
signal s(t) = cos100t + 0.1cos100tcos5t . using the identity for cosacosb
s(t) = cos100t + [tex]\frac{0.1}{2}[/tex] [cos(100+5)t + cos (100-5)t]
s(t) = cos 100t + 0.05cos ( 100+5)t + 0.05cos (100-5)t
= cos100t + 0.05cos(105)t + 0.05cos 95t
= cos 2 [tex](\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t[/tex] [ ∵cos (∅) = sin(/2 +∅ ]
= sin ( 2 [tex](\frac{50}{\pi } ) t[/tex] + /2 ) + 0.05sin ( 2 [tex](\frac{105}{2\pi } ) t + /2 )[/tex] + 0.05sin ( 2 [tex](\frac{95}{2\pi } )t + /2[/tex] )
attached is the remaining part of the solution
We need to design a logic circuit for interchanging two logic signals. The system has three inputs I1I1, I2I2, and SS as well as two outputs O1 and O2. When S is low, we should have O1 = I1 and O2 = I2. On the other hand, when S is high,we should have O1 = I2 and O2 =I1. Thus, S acts as the control input for a reversing switch. Use Karnaugh maps to obtain a minimal SOP(sum ofproduct) design. Draw the circuit.
Explanation:
Inputs and Outputs:
There are 3 inputs = I₁, I₂, and S
There are 2 outputs = O₁ and O₂
The given problem is solved in three major steps:
Step 1: Construct the Truth Table
Step 2: Obtain the logic equations using Karnaugh map
Step 3: Draw the logic circuit
Step 1: Construct the Truth Table
The given logic is
When S = 0 then O₁ = I₁ and O₂ = I₂
When S = 1 then O₁ = I₂ and O₂ = I₁
I₁ | I₂ | S | O₁ | O₂
0 | 0 | 0 | 0 | 0
0 | 0 | 1 | 0 | 0
0 | 1 | 0 | 0 | 1
0 | 1 | 1 | 1 | 0
1 | 0 | 0 | 1 | 0
1 | 0 | 1 | 0 | 1
1 | 1 | 0 | 1 | 1
1 | 1 | 1 | 1 | 1
Step 2: Obtain the logic equations using Karnaugh map
Please refer to the attached diagram where Karnaugh map is set up.
The minimal SOP representation for output O₁
[tex]$ O_1 = I_1 \bar{S} + I_2 S $[/tex]
The minimal SOP representation for output O₂
[tex]$ O_2 = I_2 \bar{S} + I_1 S $[/tex]
Step 3: Draw the logic circuit
Please refer to the attached diagram where the circuit has been drawn.
A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the inlet temperature if the gas is argon, helium, or nitrogen.
Given Information:
Inlet velocity = Vin = 25 m/s
Exit velocity = Vout = 250 m/s
Exit Temperature = Tout = 300K
Exit Pressure = Pout = 100 kPa
Required Information:
Inlet Temperature of argon = ?
Inlet Temperature of helium = ?
Inlet Temperature of nitrogen = ?
Answer:
Inlet Temperature of argon = 360K
Inlet Temperature of helium = 306K
Inlet Temperature of nitrogen = 330K
Explanation:
Recall that the energy equation is given by
[tex]$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $[/tex]
Where Cp is the specific heat constant of the gas.
Re-arranging the equation for inlet temperature
[tex]$ T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$[/tex]
For Argon Gas:
The specific heat constant of argon is given by (from ideal gas properties table)
[tex]C_p = 520 \:\: J/kg.K[/tex]
So, the inlet temperature of argon is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 119 + 300$[/tex]
[tex]$ T_{in} = 360K $[/tex]
For Helium Gas:
The specific heat constant of helium is given by (from ideal gas properties table)
[tex]C_p = 5193 \:\: J/kg.K[/tex]
So, the inlet temperature of helium is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 12 + 300$[/tex]
[tex]$ T_{in} = 306K $[/tex]
For Nitrogen Gas:
The specific heat constant of nitrogen is given by (from ideal gas properties table)
[tex]C_p = 1039 \:\: J/kg.K[/tex]
So, the inlet temperature of nitrogen is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 60 + 300$[/tex]
[tex]$ T_{in} = 330K $[/tex]
Note: Answers are rounded to the nearest whole numbers.
1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing under a constant load is to double. Use equation 11-1 as your starting point. (b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.
Answer:
(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.
Equation 11-1: F*L^(1/3) = Constant
Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483
Explanation:
(a)The Catalog rating(C)
Bearing life:[tex]L_1 = L , L_2 = 2L[/tex]
Catalog rating: [tex]C_1 = C , C_2 = ? ,[/tex]
From given equation bearing life equation,
[tex]F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)[/tex]
we Dividing eqn (2) with (1)
[tex]\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C[/tex]
The Catalog rating increased by factor of 1.26
(b) Reliability Increase from 0.9 to 0.99
[tex]R_1 = 0.9 , R_2 = 0.99[/tex]
Now calculating life adjustment factor for both value of reliability from Weibull parametres
[tex]a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}[/tex]
[tex]= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968[/tex]
Similarly
[tex]a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215[/tex]
Now calculating bearing life for each value
[tex]L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L[/tex]
Now using given ball bearing life equation and dividing each other similar to previous problem
[tex]\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C[/tex]
Catalog rating increased by factor of 0.61
Find the function f and the value of the constant a such that: 2 ∫ f(t)dt x a = 2 cos x − 1
Answer:
The function is [tex]-\sin x[/tex] and the constant of integration is [tex]C = - 1[/tex].
Explanation:
The resultant expression is equal to the sum of a constant multiplied by the integral of a given function and an integration constant. That is:
[tex]a = k\cdot \int\limits {f(x)} \, dx + C[/tex]
Where:
[tex]k[/tex] - Constant, dimensionless.
[tex]C[/tex] - Integration constant, dimensionless.
By comparing terms, [tex]k = 2[/tex], [tex]C = -1[/tex] and [tex]\int {f(x)} \, dx = \cos x[/tex]. Then, [tex]f(x)[/tex] is determined by deriving the cosine function:
[tex]f(x) = \frac{d}{dx} (\cos x)[/tex]
[tex]f(x) = -\sin x[/tex]
The function is [tex]-\sin x[/tex] and the constant of integration is [tex]C = - 1[/tex].
A three-phase line has an impedance of 0.4 j2.7 ohms per phase. The line feeds two balanced three-phase loads that are connected in parallel. The first load is absorbing 560.1kVA at 0.707 power factor lagging. The second load absorbs 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 3810.5 volts. Determine: a. The magnitude of the line voltage at the source end of the line. b. Total real and reactive power loss in the line. c. Real power and reactive power supplied at the sending end of the line.
Answer:
a. The magnitude of the line source voltage is
Vs = 4160 V
b. Total real and reactive power loss in the line is
Ploss = 12 kW
Qloss = j81 kvar
Sloss = 12 + j81 kVA
c. Real power and reactive power supplied at the sending end of the line
Ss = 540.046 + j476.95 kVA
Ps = 540.046 kW
Qs = j476.95 kvar
Explanation:
a. The magnitude of the line voltage at the source end of the line.
The voltage at the source end of the line is given by
Vs = Vload + (Total current×Zline)
Complex power of first load:
S₁ = 560.1 < cos⁻¹(0.707)
S₁ = 560.1 < 45° kVA
Complex power of second load:
S₂ = P₂×1 (unity power factor)
S₂ = 132×1
S₂ = 132 kVA
S₂ = 132 < cos⁻¹(1)
S₂ = 132 < 0° kVA
Total Complex power of load is
S = S₁ + S₂
S = 560.1 < 45° + 132 < 0°
S = 660 < 36.87° kVA
Total current is
I = S*/(3×Vload) ( * represents conjugate)
The phase voltage of load is
Vload = 3810.5/√3
Vload = 2200 V
I = 660 < -36.87°/(3×2200)
I = 100 < -36.87° A
The phase source voltage is
Vs = Vload + (Total current×Zline)
Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)
Vs = 2401.7 < 4.58° V
The magnitude of the line source voltage is
Vs = 2401.7×√3
Vs = 4160 V
b. Total real and reactive power loss in the line.
The 3-phase real power loss is given by
Ploss = 3×R×I²
Where R is the resistance of the line.
Ploss = 3×0.4×100²
Ploss = 12000 W
Ploss = 12 kW
The 3-phase reactive power loss is given by
Qloss = 3×X×I²
Where X is the reactance of the line.
Qloss = 3×j2.7×100²
Qloss = j81000 var
Qloss = j81 kvar
Sloss = Ploss + Qloss
Sloss = 12 + j81 kVA
c. Real power and reactive power supplied at the sending end of the line
The complex power at sending end of the line is
Ss = 3×Vs×I*
Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)
Ss = 540.046 + j476.95 kVA
So the sending end real power is
Ps = 540.046 kW
So the sending end reactive power is
Qs = j476.95 kvar
What's a disadvantage of highest MERV-rated filters?
1) the pressure drop across high MERV filters is less.
2) high MERV filters require at least a 240 V power supply.
3) the pressure drop across high MERV filters is significant.
Answer:
3) the pressure drop across high MERV filters is significant.
Explanation:
MERV (Minimum-Efficiency Reporting Value) is used to measure the efficiency of filter to remove particles. A filter of high MERV can filter smaller particles but this causes an increase in reduced air flow that is an increase in pressure drop. High MERV filters capture more particles causing them to get congested faster and thereby increasing pressure drop.
Excessive pressure drop can cause overheating and lead to damage of the filter. The pressure drop can be reduced by increasing the surface area of the filter.
Answer:
3) The pressure drop across high MERV filters is significant.
Explanation:
The higher the MERV filter rate is, the most efficient it will be when it comes to trapping small particles. This comes with a cost. Since the space between fibers is smaller, this translates into a higher pressure drop. This is a disadvantge since in air conditioning or ventilation systems, the higher the pressure drop, the biggest the equipment and the most expensive it is.
Using models helps scientists conduct research. How else can research using models save lives?
Answer: find the answer in the explanation.
Explanation:
Some of the methods of science are hypothesis, observations and experiments.
Scientific research and findings cut across many areas and field of life. Field like engineering, astronomy and medicine e.t.c
Using models in research is another great method in which researchers and scientists can master a process of a system and predict how it works.
Wilbur and Orville Wright wouldn't have died if the first airplane built by these two wonderful brothers was first researched by using models and simulations before embarking on a test.
Modelling in science involves calculations by using many mathematical methods and equations and also by using many laws of Physics. In this present age, many of these are computerised.
In space science, it will save lives, time and money to first model and simulate if any new thing is discovered in astronomy rather than people going to the space by risking their lives.
Also in medicine, drugs and many medical equipment cannot be tested by using people. This may lead to chaos and loss of lives.
So research using models in astronomy, medicine, engineering and many aspects of science and technology can indeed save lives.
a surveyor is trying to find the height of a hill . he/she takes a sight on the top of the hill and find that the angle of elevation is 40°. he/she move a distance of 150 metres on level ground directly away from the hill and take a second sight. from this point the angl.e of elevation is 22°. find the height of the
hill
Answer:
height ≈ 60.60 m
Explanation:
The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.
Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.
Using tangential ratio,
tan 22° = opposite/adjacent
tan 22° = h/150
h = 150 × tan 22°
h = 150 × 0.40402622583
h = 60.6039338753
height ≈ 60.60 m
Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline
Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?
Answer: Technician B is correct
Explanation: Two types of engines exist , the two stroke (example, used in chainsaws) is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the four stroke engines(eg lawnmowers) which uses four strokes, 2-strokes during compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.
While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before forcing the mixture into the cylinder and do not require a pressurized system. The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and directed to the main moving parts of crankshaft through its channels.
We can therefore say that Technician A is wrong while Technician B is correct
Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one?
A) Higher efficiency and higher effectiveness.
B) Higher efficiency but lower effectiveness.
C) Lower efficiency but higher effectiveness.
D) Lower efficiency and iower effectiveness.
E) Equal efficiency and equal effectiveness.
Answer:
D) Lower efficiency and lower effectiveness.
Explanation:
Given;
Two finned surfaces with long fins which are identical,
with difference in the convection heat transfer coefficient,
The first finned surface has a higher convection heat transfer coefficient, but gives the same heat rate as the second, which will make it (first finned surface) to have lower efficiency and lower effectiveness than the second finned surface.
Therefore, the correct option is "(D) Lower efficiency and lower effectiveness"
A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The pump is considered isentropic and the turbine isentropic efficiency is 85%. If the net power output is 100 MW calculate the thermal efficiency of the plant and the mass flow rate of steam
Answer:
0.31
126.23 kg/s
Explanation:
Given:-
- Fluid: Water
- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%
- Pump: Isentropic
- Net cycle-work output, Wnet = 100 MW
Find:-
- The thermal efficiency of the cycle
- The mass flow rate of steam
Solution:-
- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.
First process: Isentropic compression by pump
P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )
h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )
s1 = s-P1 = 0.6492 KJ/kg.K
v1 = v-P1 = 0.001010 m^3 / kg
P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )
s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )
- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:
[tex]w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}[/tex]
- From the following relation we can determine ( h2 ) as follows:
h2 = h1 + wp
h2 = 191.81 + 8.0699
h2 = 199.88 KJ/kg
Second Process: Boiler supplies heat to the fluid and vaporize
- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).
- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).
P3 = 8 MPa
T3 = ? ( assume fluid exist in the saturated vapor phase )
h3 = hg-P3 = 2758.7 KJ/kg
s3 = sg-P3 = 5.7450 KJ/kg.K
- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:
[tex]q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}[/tex]
Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).
- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.
- Under the isentropic conditions the steam exits the turbine at the following conditions:
P4 = 10 KPa
s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )
- Compute the quality of the mixture at condenser inlet by the following relation:
[tex]x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947[/tex]
- Determine the isentropic ( h4s ) at this state as follows:
[tex]h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}[/tex]
- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:
[tex]h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\[/tex]
- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.
[tex]w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}[/tex]
- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.
[tex]W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}[/tex]
Answer: The mass flow rate of the steam would be 126.23 kg/s
- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):
[tex]n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31[/tex]
Answer: The thermal efficiency of the cycle is 0.31
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given by pVn 5 constant. The initial volume is 0.1 m3, the final volume is 0.04 m3, and the final pressure is 2 bar. Determine the initial pressure, in bar, and the work for the process, in kJ, if (a) n 5 0, (b) n 5 1, (c) n 5 1.3.
Answer:
A.) P = 2bar, W = - 12kJ
B.) P = 0.8 bar, W = - 7.3 kJ
C.) P = 0.608 bar, W = - 6.4kJ
Explanation: Given that the relation between pressure and volume is
PV^n = constant.
That is, P1V1^n = P2V2^n
P1 = P2 × ( V2/V1 )^n
If the initial volume V1 = 0.1 m3,
the final volume V2 = 0.04 m3, and
the final pressure P2 = 2 bar.
A.) When n = 0
Substitute all the parameters into the formula
(V2/V1)^0 = 1
Therefore, P2 = P1 = 2 bar
Work = ∫ PdV = constant × dV
Work = 2 × 10^5 × [ 0.04 - 0.1 ]
Work = 200000 × - 0.06
Work = - 12000J
Work = - 12 kJ
B.) When n = 1
P1 = 2 × (0.04/0.1)^1
P1 = 2 × 0.4 = 0.8 bar
Work = ∫ PdV = constant × ∫dV/V
Work = P1V1 × ln ( V2/V1 )
Work = 0.8 ×10^5 × 0.1 × ln 0.4
Work = - 7330.3J
Work = -7.33 kJ
C.) When n = 1.3
P1 = 2 × (0.04/0.1)^1.3
P1 = 0.6077 bar
Work = ∫ PdV
Work = (P2V2 - P1V1)/ ( 1 - 1.3 )
Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )
Work = (8000 - 6080)/ -0.3
Work = -1920/0.3
Work = -6400 J
Work = -6.4 kJ
9. A Co has 500,000 total shares outstanding and each share is priced at 20$. B Co has 300,000 total shares outstanding and each share is priced at 40$. You have 100 shares in A Co and 200 shares in B Cos. After consolidation how many new shares you will own in consolidated AB Co?
Answer:
In consolidated AB Co 300 shares.
Explanation:
Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
[tex]T_{min} =[/tex] 26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, [tex]d_c = 0.5 mm[/tex]
Number of passes necessary for this reduction, [tex]n = \frac{\triangle d}{d_c}[/tex]
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
[tex]Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec[/tex]
minimum time required to reduce the depth of the plate by 20 mm:
[tex]T_{min} =[/tex] number of passes * Time/pass
[tex]T_{min} =[/tex] n * Time/pass
[tex]T_{min} =[/tex] 40 * 40
[tex]T_{min} =[/tex] 1600 = 26 mins 40 secs
Answer:
the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds
Explanation:
From the given information;
Assuming the tool moves 100 mm/sec
The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:
Number of passes = [tex]\dfrac{30-20}{0.5}[/tex]
Number of passes = 20
We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times
However; the time per passes is;
Time/pass = [tex]\dfrac{20*L}{velocity \ of \ the \ feed}[/tex]
where;
length L = 400mm
velocity of the feed is assumed as 100
Time/pass [tex]=\dfrac{20*400}{100}[/tex]
Time/pass = 80 sec
Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:
[tex]T_{min} = Time/pass *number of passes[/tex]
[tex]T_{min} = 20*80[/tex]
[tex]T_{min} = 1600 \ sec[/tex]
[tex]T_{min}[/tex] = 26 minutes 40 seconds
Mathematical modeling aids in technological design by simulating how.
1. A solution should be designed
2. A proposed system might behave
3. Physical models should be built
4. Designs should be used
Mathematical modeling aids in technological design by simulating how proposed system might behave. The correct option is 2.
What is mathematical modelling?Mathematical modelling describes a real world problem in mathematical terms or in the form of equations. This makes an engineer to discover new features about the problem and designer to alter his design for better function and output.
Mathematical models allow engineers and designers to understand how the proposed model and actual prototype will be produced.
Thus, the correct option is 2.
Learn more about mathematical modelling
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A vehicle is being tested using a scan tool. Technician A says that the throttle position (TP) sensor should read between 0 and 1 volt. Technician B says that a factory or factory level scan tool is needed to check for proper voltage and current of the shift solenoids. Which technician is correct?
Answer: Non of them.
Explanation:
There are many throttle position sensors voltage check. Computer reference voltage to TP sensor is about 5V. Voltage drop should be less than 0.5 v while TP sensor output to computer is about 0.65v.
When a vehicle is being tested using a scan tool, the only the throttle position (TP) sensor that should read between 0 and 1 volt is the base voltage reading. The reading according to specifications is around 0.05v.
When checking for proper voltage for the opening and closing of the throttle, the voltage rises from 1 volt to a maximum of 5 volts.
Ammeter or multimeter can be used to check for proper voltage and current of the shift solenoids.
We can therefore conclude that both technicians A and B are incorrect. This is, non of them is correct.
A heavy ball with a weight of 110 N is hung from the ceiling of a lecture hall on a 4.9-m-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.0 m/s as it passes through the lowest point.
Required:
What is the tension in the rope at that point?
Answer:T = 167.3 N
Explanation:
Given that the
Weight mg = 110 N
The mass m of the ball will be
m = 110/9.8 = 11.22 kg
As the direction of the ball’s velocity is changing, the force responsible for this is centripetal force F. And
F = mV^2/r
Where
V = 5.0 m/s
r = L = 4.9 m
m = 11.22
Substitute all these parameters into the formula
F = (11.22 × 5^2)/4.9
F = 280.6/4.9
F = 57.27 N
Tension T = F + mg
Substitute F and mg into the formula
T = 57.27 + 110
T = 167.3 N
Therefore, the tension in the rope at that point is 167.3 N
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You close the switch at t = 0. Find (a) the current in R1 and R2 at t=0, (b) the voltage across R1 after a long time. (Careful with this one.)
Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in the impeller shaft must be limited to 80 MPa, determine(a) the minimum diameter required for a solid impeller shaft.(b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm.(c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)
Answer:
a) 34 mm
b) 39 mm
c) 93.16%
Explanation:
power transmitted P = 28 kW 28000 W
angular speed N = 440 rpm
angular speed in rad/s Ω = 2[tex]\pi[/tex]N/60
Ω = (2 x 3.142 x 440)/60 = 46.08 rad/s
allowable shear stress τ = 80 MPa = 80 x [tex]10^{6}[/tex] Pa
torque T = P/Ω = 28000/46.08 = 607.64 N-m
a) for the minimum diameter of a solid shaft, we use the equation
τ[tex]d^{3}[/tex]= [tex]\frac{16T}{ \pi}[/tex]
80 x [tex]10^{6}[/tex] x [tex]d^{3}[/tex] = [tex]\frac{16*607.64}{3.142}[/tex] = 3094.28
[tex]d^{3}[/tex] = 3094.28/(80 x [tex]10^{6}[/tex]) = 0.0000386785
d = [tex]\sqrt[3]{0.0000386785}[/tex] ≅ 0.034 m = 34 mm
b) For a hollow shaft with outside diameter D = 40 mm = 0.04 m
we use the equation,
T = [tex]\frac{16}{\pi }[/tex] x τ x [tex]\frac{D^{4} - d^{4}}{D^{4} }[/tex]
where d is the internal diameter of the pipe
607.64 = [tex]\frac{16}{3.142}[/tex] x 80 x [tex]10^{6}[/tex] x [tex]\frac{0.04^{4} - d^{4}}{0.04^{4} }[/tex]
3.82 x [tex]10^{-12}[/tex] = [tex]0.04^{4} - d^{4}[/tex]
[tex]d^{4}[/tex] = [tex]\sqrt[4]{2.56*10^{-6} }[/tex]
d = 0.039 m = 39 mm
c) we assume weight is proportional to cross-sectional area
for solid shaft,
area = [tex]\pi r^{2}[/tex]
r = diameter/2 = 34/2 = 17 mm
area = 3.142 x [tex]17^{2}[/tex] = 907.92 mm^2
for hollow shaft, radius is also gotten as before
external area = [tex]\pi r^{2}[/tex] = 3.142 x [tex]20^{2}[/tex] = 1256.64 mm^2
internal diameter = [tex]\pi r^{2}[/tex] = 3.142 x [tex]19.5^{2}[/tex] = 1194.59 mm^2
true area of hollow shaft = external area minus internal area
area = 1256.64 - 1194.59 = 62.05 mm^2
material weight saved is proportional to 907.92 - 62.05 = 845.87 mm^2
percentage weight saved is proportional to 845.87/907.92 x 100%
= 93.15%
Steam is contained in a closed rigid container which has a volume of 2 initially the the pressure and the temperature is the remeraturedrops as a result of heat transfer to the surroundings. Determine
a) the temperature at which condensation first occurs, in °C,
b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
c) What is the volume, in m3, occupied by saturated liquid at the final state?
The given question is incomplete. The complete question is as follows.
Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine
(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],
(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?
Explanation:
Using the property tables
[tex]T_{1} = 500^{o}C[/tex], [tex]P_{1}[/tex] = 10 bar
[tex]v_{1} = 0.354 m^{3}/kg[/tex]
(a) During the process, specific volume remains constant.
[tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]
T = [tex](150 - 160)^{o}C[/tex]
Using inter-polation we get,
T = [tex]154.71^{o}C[/tex]
The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].
(b) When the system will reach at state 3 according to the table at 0.5 bar then
[tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]
[tex]v_{g} = 3.24 m^{3} kg[/tex]
Let us assume "x" be the gravity if stream
[tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]
[tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]
= [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]
= 0.109
At state 3, the fraction of total mass condensed is as follows.
[tex](1 - x_{5})[/tex] = 1 - 0.109
= 0.891
The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.
(c) Hence, total mass of the system is calculated as follows.
m = [tex]\frac{v}{v_{1}}[/tex]
= [tex]\frac{1}{0.354}[/tex]
= 2.825 kg
Therefore, at final state the total volume occupied by saturated liquid is as follows.
[tex]v_{ws} = m \times v_{f}[/tex]
= [tex]2.825 \times 0.00103[/tex]
= [tex]2.9 \times 10^{-3} m^{3}[/tex]
The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].
A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch W, what tension T in the cable will be required?
The 600-kg trunk has a center
of gravity at G. The felling notch at O is sufficiently large so that the resisting moment there is negligible. Enter your answer in (N) - numerical value only
Answer:
The correct answer will be "400.4 N". The further explanation is given below.
Explanation:
The given values are:
Mass of truck,
m = 600 kg
g = 9.8 m/s²
On equating torques at the point O,
⇒ [tex]T\times Cos(10+5)\times (1.3+4)=mg\times Sin(5)\times 4[/tex]
So that,
On putting the values, we get
⇒ [tex]T\times Cos(15^{\circ})\times 5.3=600\times 9.8\times Sin(5^{\circ})\times 4[/tex]
⇒ [tex]T=400.4 \ N[/tex]
A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the rotor at synchronous speed. (b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses. (c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.
Given Information:
Frequency = f = 60 Hz
Complex rated power = G = 100 MVA
Intertia constant = H = 8 MJ/MVA
Mechanical power = Pmech = 80 MW
Electrical power = Pelec = 50 MW
Number of poles = P = 4
No. of cycles = 10
Required Information:
(a) stored energy = ?
(b) rotor acceleration = ?
(c) change in torque angle = ?
(c) rotor speed = ?
Answer:
(a) stored energy = 800 Mj
(b) rotor acceleration = 337.46 elec deg/s²
(c) change in torque angle (in elec deg) = 6.75 elec deg
(c) change in torque angle (in rmp/s) = 28.12 rpm/s
(c) rotor speed = 1505.62 rpm
Explanation:
(a) Find the stored energy in the rotor at synchronous speed.
The stored energy is given by
[tex]E = G \times H[/tex]
Where G represents complex rated power and H is the inertia constant of turbo-generator.
[tex]E = 100 \times 8 \\\\E = 800 \: MJ[/tex]
(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.
The rotor acceleration is given by
[tex]$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2} $[/tex]
Where M is given by
[tex]$ M = \frac{E}{180 \times f} $[/tex]
[tex]$ M = \frac{800}{180 \times 50} $[/tex]
[tex]M = 0.0889 \: MJ \cdot s/ elec \: \: deg[/tex]
So, the rotor acceleration is
[tex]$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2} $[/tex]
[tex]$ 30 = 0.0889 \frac{d^2 \delta}{dt^2} $[/tex]
[tex]$ \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889} $[/tex]
[tex]$ \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $[/tex]
(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.
The change in torque angle is given by
[tex]$ \Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $[/tex]
Where t is given by
[tex]1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec[/tex]
So,
[tex]$ \Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $[/tex]
[tex]$ \Delta \delta = 6.75 \: elec \: deg[/tex]
The change in torque in rpm/s is given by
[tex]$ \Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ } $[/tex]
[tex]$ \Delta \delta =28.12 \: \: rpm/s $[/tex]
The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by
[tex]$ Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t $[/tex]
Where P is the number of poles of the turbo-generator.
[tex]$ Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2 $[/tex]
[tex]$ Rotor \: speed = 1500 + 5.62 $[/tex]
[tex]$ Rotor \: speed = 1505.62 \:\: rpm[/tex]
With a very precise volumetric measuring device, the volume of a liquid sample is determined to be 6.321 L (liters). Three students are asked to determine the volume of the same liquid sample using a less precise measuring instrument. How do you evaluate the following work of each student with regards to precision, and accuracy
Students
Trials A B C
1 6.35L 6.31L 6.38L
2 6.32L 6.31 L 6.32L
3 6.33L 6.32L 6.36L
4 6.36L 6.35L 6.36L
Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:
[tex]E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\[/tex]
- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:
[tex]Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\[/tex]
- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .
[tex]P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\[/tex]
- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile. (2 marks) d) Find the speed of the automobile at the time it was overtaken by the police car. (2 marks)
Answer:
A.) Time = 13.75 seconds
B.) Total distance = 339 m
C.) V = 11.18 m/s
D.) V = 10.2 m/s
Explanation: Given that the automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone.
Then,
Initial velocity U of the motorist = 15.65m/s
acceleration a = - 3.05 m/s^2
Initial velocity u of the police man = 11.18 m/s
Acceleration a = 1.96 m/s^2
The police will overtake at distance S as the motorist decelerate and come to rest.
Where V = 0 and a = negative
While the police accelerate.
Using 2nd equation of motion for the motorist and the police
S = ut + 1/2at^2
Since the distance S covered will be the same, so
15.65t - 1/2×3.05t^2 = 11.18t +1/2×1.96t^2
Solve for t by collecting the like terms
15.56t - 1.525t^2 = 11.18t + 0.98t^2
15.56t - 11.18t = 0.98t^2 + 1.525t^2
4.38t = 2.505t^2
t = 4.38/2.505
t = 1.75 seconds approximately
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit.
Therefore, the total time required for the police car to overtake the automobile will be:
12 + 1.75 = 13.75 seconds
B.) Using the same formula
S = ut + 1/2at^2
Where S = total distance travelled
Substitutes t into the formula
S = 11.18(13.75) + 1/2 × 1.96 (13.75)^2
S = 153.725 + 185.28
S = 339 m approximately
C.) The speed of the police car at the time it overtakes the automobile will be constant = 11.18 m/s
D.) Using first equation of motion
V = U - at
Since the motorist is decelerating
V = 15.65 - 3.05 × 1.75
V = 15.65 - 5.338
V = 10.22 m/s
Therefore, the speed of the automobile at the time it was overtaken by the police car is 10.2 m/ s approximately
Assuming the temperature of the system is 20 C the atmospheric pressure is 0.998 atm, and the total volume collected in the balloon is 0.28 L, calculate the number of moles of H2 collected in the ballot. This is the actual yield. Use the dalton's law of partial pressure to determine the partial pressure of H2 in the balloon. The vapor pressure of water is 20 Cis 17.5 mmHg
Answer:
The number of moles of H₂ is 0.01135 moles
The partial pressure of H₂ in the balloon is 0.975 atm
Explanation:
From Dalton's law of partial pressure, we have;
Total pressure = Pressure of H₂ + Pressure of the water vapor
The vapor pressure of water at 20°C = 17.5 mmHg = 0.0230225 atm
For equilibrium, pressure of the balloon = Surrounding pressure = Atmospheric pressure
∴ Pressure of H₂ + Pressure of the water vapor = Atmospheric pressure
Partial pressure of H₂ in the balloon, P = 0.998 - 0.0230225 = 0.9749775 ≈ 0.975 atm
0.975 atm = 98789.595 Pa
Volume of balloon, V = 0.28 L
n = PV/(RT)
Where:
R = Universal gas constant = 0.08205 L·atm/(mol·K)
T = Temperature = 20°C = 293.15 K
∴ n = 0.975 * 0.28/ (0.08205 * 293.15) = 0.01135 moles.
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!