Answer:
Rated power = 1345.66 W/m²
Mechanical power developed = 3169035.1875 W
Explanation:
Wind speed, V = 13 m/s
Coefficient of performance of turbine, [tex]C_p[/tex] = 0.3
Rotor diameter, d = 100 m
or
Radius = 50 m
Air density, ρ = 1.225 kg/m³
Now,
Rated power = [tex]\frac{1}{2}\rho V^3[/tex]
or
Rated power = [tex]\frac{1}{2}\times1.225\times13^3[/tex]
or
Rated power = 1345.66 W/m²
b) Mechanical power developed = [tex]\frac{1}{2}\rho AV^3C_p[/tex]
Here, A is the area of the rotor
or
A = π × 50²
thus,
Mechanical power developed = [tex]\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3[/tex]
or
Mechanical power developed = 3169035.1875 W
The bars of the truss each have a cross-sectional area of 1.25 in2. If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss.
Answer:
P=25000lbf
Explanation:
For this problem we will use the equation that relates, the effort, the area and the force for an element under normal stress.
σ=P/A
σ=stress=20kSI=20 000 lbf/in ^2
P=force
A=area
solving for P
P=Aσ
P=(20 000 lbf/in ^2)(1.25in^2)
P=25000lbf
Answer:
The maximum magnitude P of the loads that can be applied to the truss = 25,000 Pounds or 111,205.5 Newtons.
Explanation:
In order to calculate the maximum load P, we will make use of the formula: Maximum average stress (20 ksi) = maximum load P ÷ cross-sectional area (1.25 in²)
Make P (the maximum load) the subject of the formula: P = 20 ksi × 1.25 in².
Before moving further, we have to convert the average normal stress (in ksi) to an appropriate unit: The average normal stress = 20 ksi = 20 kip per square inch (kip/in²)
But 1 kip = 1000 Pounds (i.e., 1000 lb)
Therefore, 20 ksi = 20,000 Pounds/in².
Therefore, P (maximum load) = 20,000 pounds/in² × 1.25 in² = 25,000 Pounds = 111,205.5 Newtons (because 1 Pound = 4.44822 Newtons).
Select the most accurate response. Refrigerators and heat pumps are both systems which transfer energy from low-temperature reservoirs to high-temperature reservoirs, against the natural direction of heat transfer. The purpose of a refrigerator is to keep an enclosed space cool. The purpose of a heat pump is: Select one: a. to remove heat from a colder space b. to remove heat from a warmer space c. to supply heat to a colder space d. to supply heat to a warmer space
Answer:
d. to supply heat to a warmer space
Explanation:
Let's analyze the options
(a) to remove heat from a colder space is the purpose of a refrigerator. Note that heat pump also do this
(b) to remove heat from a warmer space. This option has no sense, both refrigerators and heat pumps supply heat to a warmer space
(c) to supply heat to a colder space. This option has no sense, both refrigerators and heat pumps remove heat from a colder space
(d) to supply heat to a warmer space. This is the purpose of a heat pump
The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in Appendix A to obtain a conversion
Answer:
The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)
Explanation:
Since the given volume flow rate is gallons per minute.
We know that 1 gallon = 3.785 liters and
1 minute = 60 seconds
Let the flow rate be [tex]Q\frac{gallons}{minute}[/tex]
Now replacing the gallon and the minute by the above values we get
[tex]Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}[/tex]
Thus [tex]Q'=0.631Q\frac{liters}{second}[/tex]
Now since we know that 1 liter = [tex]0.0353ft^{3}[/tex]
Using this in above relation we get
[tex]Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q[/tex]
From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.
A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 MN. Will the sample experience plastic deformation? You must justify your answer.
Answer:Yes,266.66 MPa
Explanation:
Given
Yield stress of material =140 MPa
Cross-section of [tex]300\times 100 mm^2[/tex]
Force(F)=8 MN
Therefore stress due to this Force([tex]\sigma [/tex])
[tex]\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}[/tex]
[tex]\sigma =266.66 \times 10^{6} Pa[/tex]
[tex]\sigma =266.66 MPa[/tex]
Since induced stress is greater than Yield stress therefore Plastic deformation occurs
One way to lower NOx emissions in diesel engines is to add water to the fuel. Since water is not mixable with diesel fuel, it has to been emulsified by a surfactant. Explain how it works and why.
After the emulsion is made, fuel droplet containing water droplets increase its temperature inside the engine, because of that, water droplets explode causing water rapid evaporation that break down fuel droplet, in consequence, smaller fuel droplets are generated. This makes the temperature of the flame during the combustion decrease, and that is why the reduction of NOx is accomplished by this method .
The capacitor is initially discharged, and the battery is at a positive voltage. Which correctly describes what happens after the switch has remained closed for a long time? Note: The battery is ideal (no internal resistance). The wires are ideal (no connecting resistance). The resistance of the bulb does not change as current flows through it. The switch is ideal; it has no resistance when closed, and infinite resistance when open. The capacitor is ideal (no internal resistance or inductance).
Answer:
The lamp will remain off.
Explanation:
I assume this is a simple series circuit with a battery, a switch, an incandescent lamp and a capacitor.
Since the circuit is of continuous current (because it uses a battery), after the switch has remained closed for a long time the capacitor will be fully charged. Being fully charged, no continuous current will flow through the capacitor, and since it is in series with the rest of the circuit there will be no current anywhere, so the lamp will remain off.
Calculate the time taken to completely empty aswimming pool 15
m long and 9 m wide through an opening at thebottom as shown in the
fig. The swimming pool holds water to depth1.5 m.Opening area is
0.3 sq.m and Cd=0.62.
Answer:
Time needed to empty the pool is 401.35 seconds.
Explanation:
The exit velocity of the water from the orifice is obtained from the Torricelli's law as
[tex]V_{exit}=\sqrt{2gh}[/tex]
where
'h' is the head under which the flow of water occurs
Thus the theoretical discharge through the orifice equals
[tex]Q_{th}=A_{orifice}\times \sqrt{2gh}[/tex]
Now we know that
[tex]C_{d}=\frac{Q_{act}}{Q_{th}}[/tex]
Thus using this relation we obtain
[tex]Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}[/tex]
Now we know by definition of discharge
[tex]Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}[/tex]
Using the above relations we obtain
[tex]Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\[/tex]
The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0
Solving for 't' we get
[tex]\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.[/tex]
Calculate the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, the centers of which are separated by a distance of 2.9 nm.
Calculating the force of attraction between a cation and an anion using Coulomb's Law involves converting the given distance into meters, using the elementary charge for valences, and then applying the formula.
Explanation:The question is about calculating the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, separated by a distance of 2.9 nm. To find this, we use Coulomb's Law, which is given by the equation F = k * |q1*q2| / r2, where F is the force between the charges, k is Coulomb's constant (8.9875517873681764 × 109 N m2 C-2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges. However, given the units of distance as nanometers (nm) and charge in terms of electron charge, we first convert the distance to meters (1 nm = 1 × 10-9 meters) and use the elementary charge (1.6022 x 10-19 C) for the charges. With a +2 valence for the cation and a -1 valence for the anion, the charges become +2e and -1e, respectively. The distance r is 2.9 nm or 2.9 × 10-9 meters. Plugging these values into Coulomb's formula, the attraction force can be calculated, taking into account that the force will be negative, signifying attraction.
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a pressure of 101 kPa on top of the 0.5-m-diameter piston. The gage pressure of the air inside the cylinder is 1.2 kPa. The local acceleration of gravity is g = 9.81 m/s2 . Subsequently, a weight is placed on top of the piston causing the piston to fall until reaching a new static equilibrium position. At this position, the gage pressure of the air inside the cylinder is 2.8 kPa. Determine (a) the mass of the piston, in kg, and (b) the mass of the added weight, in kg
Answer:
a) 24 kg
b) 32 kg
Explanation:
The gauge pressure is of the gas is equal to the weight of the piston divided by its area:
p = P / A
p = m * g / (π/4 * d^2)
Rearranging
p * (π/4 * d^2) = m * g
m = p * (π/4 * d^2) / g
m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg
After the weight is added the gauge pressure is 2.8kPa
The mass of piston plus addded weight is
m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg
56 - 24 = 32 kg
The mass of the added weight is 32 kg.
An exercise room has six weight-lifting machines that have no motors and seven treadmills, each equipped with a 2.5-hp (shaft output) motor. The motors operate at an average load factor of 0.7, at which their efficiency is 0.77. During peak evening hours, all 13 pieces of exercising equipment are used continuously, and there are also two people doing light exercises while waiting in line for one piece of the equipment. Assuming the average rate of heat dissipation from people in an exercise room is 740 W, determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions.
Answer:
25.4 kW
Explanation:
There are 15 people doing exercises, each will dissipate 740 W of heat, so they will disspiate a total of p = 15*740 = 11100 W = 11.1 kW
There are 7 treadmills, each has a 2.5 hp motor (1.86 kW) running at a load factor of 0.7 with an efficiency of 0.77. So their total power would be p = 7*1.86*0.77/0.7 =14.3 kW
So the total heat dissipated would be 11.1 + 14.3 = 25.4 kW.
The rate of heat gain from people and equipment at peak load is approximately 14.955 kW, including treadmill and people's heat dissipation.
To calculate the rate of heat gain of the exercise room from people and equipment at peak load conditions, we need to consider the heat dissipated by both the exercising equipment and the people in the room.
1. Heat dissipation from exercising equipment:
- For the treadmills: Each treadmill has a 2.5-hp motor operating at 0.7 load factor and 0.77 efficiency. So, the power consumed by each treadmill is [tex]\( P_{\text{treadmill}} = 2.5 \, \text{hp} \times 0.7 \times 0.77 = 1.925 \, \text{kW} \).[/tex]
- For the weight-lifting machines: Since they have no motors, they do not contribute to heat dissipation.
2. Heat dissipation from people:
- There are two people doing light exercises. Assuming each person dissipates heat at a rate of 740 W, the total heat dissipated by people is [tex]\( P_{\text{people}} = 2 \times 740 \, \text{W} = 1480 \, \text{W} \).[/tex]
Now, we can calculate the total rate of heat gain of the exercise room:
[tex]\[ \text{Total heat gain} = P_{\text{treadmill}} \times \text{number of treadmills} + P_{\text{people}} \][/tex]
Given there are 7 treadmills, we have:
[tex]\[ \text{Total heat gain} = 1.925 \, \text{kW} \times 7 + 1480 \, \text{W} \]\[ \text{Total heat gain} = 13.475 \, \text{kW} + 1480 \, \text{W} \]\[ \text{Total heat gain} \approx 13.475 \, \text{kW} + 1.48 \, \text{kW} \]\[ \text{Total heat gain} \approx 14.955 \, \text{kW} \][/tex]
So, the rate of heat gain of the exercise room from people and equipment at peak load conditions is approximately 14.955 kW.
Which component stores energy as potential energy in a mechanical system ?
Answer:
Spring
Explanation:
We know that energy associated with motion is known as kinetic energy and the energy associated with position or elevation is known as potential energy.
A ball on floor which on the rest condition have zero potential energy but on the other hand a ball at a height h from the floor have mgh potential energy.
Spring is the mechanical component which stored potential energy.
A heat engine receives heat from a heat source at 1453 C and has a thermal efficiency of 43 percent. The heat engine does maximum work equal to 539 kJ. Determine: a) the heat supplied to the heat engine by the heat source (kJ), b) the heat rejected to the heat sink (kJ), and c) the temperature of the heat sink (C).
Answer:
a) 1253 kJ
b) 714 kJ
c) 946 C
Explanation:
The thermal efficiency is given by this equation
η = L/Q1
Where
η: thermal efficiency
L: useful work
Q1: heat taken from the heat source
Rearranging:
Q1 = L/η
Replacing
Q1 = 539 / 0.43 = 1253 kJ
The first law of thermodynamics states that:
Q = L + ΔU
For a machine working in cycles ΔU is zero between homologous parts of the cycle.
Also we must remember that we count heat entering the system as positiv and heat leaving as negative.
We split the heat on the part that enters and the part that leaves.
Q1 + Q2 = L + 0
Q2 = L - Q1
Q2 = 539 - 1253 = -714 kJ
TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:
η = 1 - T2/T1
T2/T1 = 1 - η
T2 = (1 - η) * T1
The temperatures must be given in absolute scale (1453 C = 1180 K)
T2 = (1 - 0.43) * 1180 = 673 K
673 K = 946 C
The sticker inside the door of my car says that the tire pressure should be 32 psig (322 kPa) when the tire is cold. Before a road trip, I fill the tire to this pressure on a cold morning when the temperature is 15 °C, and then head out towards Las Vegas. When I make a rest stop in Barstow, it is now quite warm out, and the air in my tires has also warmed up from friction during the long drive. So, the air in the tires is now 60 °C. Assuming my tires don't leak or expand (volume is constant), what is the expected tire pressure at this rest stop?
Answer:
37 psi
Explanation:
For ideal gases this equation applies:
p1*V1/T1 = p2*V2/T2
Since we are assuming volume remains constant:
V2 = V1
p1/T1 = p2/T2
p2 = p1*T2/T1
The temperatures must be in absolute scale.
T1 = 15 + 273 = 288 K
T2 = 60 + 273 = 333 K
Then:
p2 = 32 * 333 / 288 = 37 psi
What system of units is superior, SI, BG, or EE? Why?
Answer:
SI unit of system is superior.
Explanation:
The metric system is used internationally for measurement of unit.This metric system of unit is easier to understand .
Metric system of unit also known as international system of unit.The international system of unit is also represent in short form like SI unit.This unit system is internationally accepted.
So SI unit of system is superior.
A shaft is to transmit 3.5 kW power while rotating at 350 rpm. If the shaft is made of plain carbon steel with 100 MPa yield strength, calculate: a) Torque being transmitted by the shaft b) Diameter of the shaft
Answer:
a)T=95.5414 N.m
b)d=21.35 mm
Explanation:
Given that
P=3.5 KW
Speed N=350 RPM
Yield strength= 100 MPa
So Shear strength = 0.5 x 100 =50 MPa
We know that
[tex]P=\dfrac{2\pi NT}{60}[/tex]
Where N is the speed ,T is the torque and P is the power.
Now by putting the values
[tex]P=\dfrac{2\pi NT}{60}[/tex]
[tex]3500=\dfrac{2\pi 350T}{60}[/tex]
T=95.5414 N.m
T=95,541.4 N.mm
We also know that
[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]
d is the diameter of shaft
[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]
[tex]50=\dfrac{16\times 95541.4}{\pi d^3}[/tex]
d=21.35 mm
Torque,T=95.5414 N.m
Diameter ,d=21.35 mm
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A residential heat pump has a coefficient of performance of 1.49 How much heating effect, in kJ/h, will result when 4 kW is supplied to this heat pump?
Answer:
21.456 kJ/h
Explanation:
See the figure attached. In this case
[tex]W_{cycle} = 4 kW [/tex]
[tex]Q_{out} = \text{heating effect}[/tex]
Coefficient of performance in heat pump is defined by
[tex]COP = \frac{Q_{out}}{W_{cycle}} [/tex]
[tex]Q_{out} =COP*W_{cycle} [/tex]
[tex]Q_{out} =1.49*4 \, W [/tex]
[tex]Q_{out} = 5.96 \, W [/tex]
Now it is necessary to change units, remember that Watt (W) is defined as J/s
[tex]Q_{out} = 5.96 \frac{J}{s} \frac{3600s}{1 h} \frac{1 kJ}{1000 J}[/tex]
[tex]Q_{out} = 21.456 \frac{kJ}{h} [/tex]
Select the most accurate statement. A diffuser converts some of a fluid's _(i)____ to ___(ii)___. Select one: a. (i) pressure, (ii) workb. (i) work, (ii) pressurec. (i) kinetic energy, (ii) heat d. (i) heat, (ii) kinetic energye. (i) enthalpy, (ii) kinetic energy f. (i) kinetic energy, (ii) enthalpy
Answer:
1.Kinetic energy 2.Enthalpy
Explanation:
We know that diffuser is a device which convert kinetic energy of fluids into pressure energy of fluids,on the other hand nozzle is used to convert the pressure energy of fluids into kinetic energy of fluids. If we want exit velocity of fluid is too high then we use nozzle and if we wand exit velocity of fluid is low then we use diffuser.
We know that enthalpy is also related with pressure so we can say that ,diffuser covert kinetic energy into enthalpy.
Enthalpy h= u + p.v
Where u is the internal energy ,p is the pressure and v is volume.
What is the lehr and what purpose does it serve?
Explanation:
Step1
Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.
Step2
Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product. So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.
Please explain what is the difference between engineering stress and true stress in a tensile test?
Answer:
Explanation:
Engineering Stress is defined as Load applied to the original cross-sectional area which we have taken in the start.
True stress is defined as the load divided by area of cross-section of specimen at that instant.
Engineering stress and true stress can be expressed by relation
[tex]\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )[/tex]
Where
[tex]\sigma _T=True\ stress[/tex]
[tex]\sigma _E=Engineering\ stress[/tex]
[tex]\epsilon _E=Engineering\ Strain[/tex]
Consider a nuclear power plant that produces 1200 MW of power and has a conversion efficiency of 34 percent (that is, for each unit of fuel energy used, the plant produces 0.34 units of electrical energy. Assuming continuous operation, determine the amount of nuclear fuel consumed by the plant per year.
Answer with Explanation:
The relation between power and energy is
[tex]Energy=Power\times Time[/tex]
Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals
[tex]E=1200\times 10^{6}\times 3600\times 24\times 365=3.784\times 10^{16}[/tex]
Now from the energy mass equivalence we have
[tex]E=mass\times c^2[/tex]
where
'c' is the speed of light in free space
Thus equating both the above values we get
[tex]3.784\times 10^{16}=mass\times (3\times 10^{8})^{2}\\\\\therefore mass=\frac{3.784\times 10^{16}}{9\times 10^{16}}=0.42kg[/tex]
Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is
[tex]mass_{req}=\frac{mass}{\eta }=\frac{0.43}{0.34}=1.235[/tex]
Thus 1.235 kg of nuclear fuel is reuired for operation.
Which of the following statements is true about machines? a)- Machines are structures made of one or more movable parts. b)- Machines have a least one multiforce member as part. c)- Machines transmit forces or alter the effect . d)- Machines, unlike frames, can’t be constrained
Answer:
Option c is True
Explanation:
a)- Machines are structures made of one or more movable parts. is false statement because Machines can be made without any movable part as in simple machines like inclined plains, wedges, screw etc.
b) Machines have a least one multi force member as part. is false statement since it is not a compulsory condition.
c) Machines transmit force or alter the effect. This is a true statement as machine are made in the first place to transmit forces and alter the effect.
d) Machines can be Contrained.
Tensile strength is a measure of a material’s resistance to being pulled apart. a) True b) False
Answer:
The given statement is correct.
Explanation:
When we load a material axially stresses are developed in the material. these stresses arise internally to keep the object in equilibrium.
When a tensile load is applied to an material since the nature of the force is to cause elongation in the material, axial strain is developed in the material or we can say that any arbitrary point in the material undergoes a displacement in the direction of the applied load. The stresses that are developed in the material tend to tear the material apart.
Strength of a material by definition is it's ability to resist deformation, thus tensile strength can be defined as the resistance of the material to tensile strains which have the tendency to tear the material apart.
What is the function of air preheater?
Answer and Explanation:
AIR PREHEAT ER :
Air preheater is a device which basically used for heating purpose of air.Air preheating is done before any process as like combustion.Air preheating is done to increase the efficiency of the process.These are also known as heat exchangerThey are mainly used in boilers for heating purpose The energy saving is dependent on air preheatingThe principal component in glass manufacturing is_____
Answer:
Silica is the principal component in glass.
Explanation:
Step1
Glass is the amorphous solid and transparent. Glass products have many of the shapes and design that are available in market.
Step2
Natural quartz is the primary source of glass in sand. Silica is the principal component in approximately all glass. Lime stone, soda ash and aluminum oxide are added in the galas for desired properties depending upon application. So, silica is the principal component in glass.
The primary heat transfer mechanism that sears my skin if I fall in a campfire is: a)- Syncrotonization b)-Radiation c)- Conduction d)-Convection
Answer:
c)Conduction
Explanation:
As we know that conduction heat transfer is take place due to movement of electron.When we fall on the campfire then the primary heat transfer mechanism is conduction because .
Convection heat transfer taker place due to motion of fluid.
Radiation heat transfer dominates at very high temperature but in campfire the temperature is not too high.
So option C is correct.
A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such that its position is observed to be s = (15t^3 - 3t) mm, where t is measured in seconds. Determine (a) the particle's displacement from t = 2 s to t = 4 s, and (b) the velocity and acceleration of the particle when t = 5 s.
Answer:
a)Δs = 834 mm
b)V=1122 mm/s
[tex]a=450\ mm/s^2[/tex]
Explanation:
Given that
[tex]s = 15t^3 - 3t\ mm[/tex]
a)
When t= 2 s
[tex]s = 15t^3 - 3t\ mm[/tex]
[tex]s = 15\times 2^3 - 3\times 2\ mm[/tex]
s= 114 mm
At t= 4 s
[tex]s = 15t^3 - 3t\ mm[/tex]
[tex]s = 15\times 4^3- 3\times 4\ mm[/tex]
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
[tex]V=\dfrac{ds}{dt}[/tex]
[tex]\dfrac{ds}{dt}=45t^2-3[/tex]
At t= 5 s
[tex]V=45t^2-3[/tex]
[tex]V=45\times 5^2-3[/tex]
V=1122 mm/s
We know that acceleration a
[tex]a=\dfrac{d^2s}{dt^2}[/tex]
[tex]\dfrac{d^2s}{dt^2}=90t[/tex]
a= 90 t
a = 90 x 5
[tex]a=450\ mm/s^2[/tex]
A bridge has been constructed between the mainland and
anisland. The total cost (excluding toll) to travel across
thebridge is expressed as C=50+0.5V, where V is the number of
veh/dayand C is the cost/vehicle in cents. The demand for
travelacross the bridge is V=2500-10C.
(a)Determine the volume of traffic across the bridge.
(b)If a toll of 25 cents is added, what is the volume
acrossthe bridge?
Answer:
a) 333,33 b)291,67
Explanation:
you have both equations :
C=50+0.5V and V=2500-10C
and you want to know the variable V so, you can calculate V in function of C, and you have already clear the variable "C", then you replace for (a):
V=2500-10CV=2500-10(50+0.5V)V=2500-500-5VV+5V=2500-5006V=2000V=2000/6V=333,33for b) they tell you that you increase your constant of the equiation C=50+0.5V (remember that the constant is the one alone, wihout any variable, in this case "50") increase in 25 so, your equiation for this point of C is C=75+0.5V, then you do the same:
V=2500-10CV=2500-10(75+0.5V)V=2500-750-5VV+5V=2500-7506V=1750V=1750/6V=291,67It's a system of two equations and two variables, wich gives you a compatible define system, it gives you only one solution.
I hope it helps you.
To determine the volume of traffic across the bridge, we can solve an equation relating cost and traffic volume. The volume of traffic across the bridge is approximately 333 vehicles per day. If a toll of 25 cents is added, the volume of traffic across the bridge is approximately 292 vehicles per day.
Explanation:To determine the volume of traffic across the bridge, we need to solve the equation for V. The equation relating cost and volume of traffic is given by V = 2500 - 10C. Substituting the cost function C = 50 + 0.5V into the equation, we get V = 2500 - 10(50 + 0.5V). Simplifying the equation, we find V = 2500 - 500 - 5V. Combining like terms, we get 6V = 2000, which yields V = 333.33. Therefore, the volume of traffic across the bridge is approximately 333 vehicles per day.
To find the volume across the bridge if a toll of 25 cents is added, we can modify the cost function to be C = 50 + 0.5V + 25 cents. We again substitute this into the equation V = 2500 - 10C and solve for V. Substituting in the new cost function, we get V = 2500 - 10(50 + 0.5V + 25). Simplifying the equation, we find V = 2500 - 10(75 + 0.5V). Continuing to simplify, we have V = 2500 - 750 - 5V. Combining like terms, we get 6V = 1750, which yields V = 291.67. Therefore, the volume across the bridge with the toll of 25 cents is approximately 292 vehicles per day.
A U tube manometer has water poured into the left side and oil (density is 790 kg/m^3) poured into to the right side. The water in left side is measured to be 70 cm high, while the right side contains both oil and water with oil height 4 times as high as water height. Determine the height of oil and height of water in right side of U tube.
Answer:
1) Height of oil in right limb = 67.31 cm
2) Height of water in the right limb = 16.83 cm
Explanation:
The U-tube manometer is shown in the attached figure
Foe equilibrium The pressure at the bottom of the U tube should be same
Let the height of the water in the left limb of the manometer be [tex]h_L[/tex]
Thus the pressure at the bottom is found using the equation of pressure statics as
[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{L}.............(i)[/tex]
Similarly for the liquid in the right limb the pressure at the bottom is the sum of the oil column and the water column
Thus we can write
[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}...........(ii)[/tex]
Equating the equations 'i' and 'ii' we get
[tex]P_{atm}+\rho _{water}\times g\times h_{L}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}\\\\\rho _{water}\times g\times 0.7=\rho _{oil}\times 4h\times g+\rho _{water}\times h\times g\\\\\therefore h_{}=\frac{0.7\times \rho _{water}}{4\times \rho _{oil}+\rho _{water}}\\\\h_{}=\frac{0.7\times 1000}{4\times 790+1000}=16.83cm[/tex]
Thus the height of oil is [tex]4\times 16.83=67.31cm[/tex] andthe height of water in the right limb is 16.83 cm.
Which two forms of energy are usually converted directly from solar radiation for houses in temperate climates. a. Thermal and Electrical b. Chemical and Nuclear c. Mechanical and Electrical
Answer:
The correct answer is option 'a': Thermal and Electrical energy
Explanation:
We know that the solar radiation that we receive from sun provides us heat. Thus the solar radiation is a natural source of thermal energy which can be utilized in solar cooker's to cook food or to warm water to generate steam in Solar thermal power plants. The steam generated is used to drive a turbine and hence produce electricity.
As we know that solar radiation is converted into directly usable forms of energy by the solar panels. The solar panels work on the principle of photo-electric effect in which light energy is directly converted into electrical energy to run our electrical devices at home such as light bulbs, fans, e.t.c. An excellent application of this principle is the international space station that orbits the earth and is fully powered by solar energy.
What is 1000J in Btu?
Answer:
0.948 Btu
Explanation:
1 Btu = 1055 J so [tex]\frac{1000}{1055}[/tex] = 0.948 Btu
1 BTU = 1055.06 joules **
(1000 J) x (1 BTU / 1055.06 J) = 0.9478 BTU
** Note: You'll have to accept my conversion factor, since I went and looked it up and you didn't.