Answer:
[tex]a=0.0937 \ m/s^2[/tex]
Explanation:
Given that
Angular velocity (ω)= 0.6 rad/s
Angular acceleration (α)= 0.3 [tex]rad/s^2[/tex]
Radius (r)= 0.2 m
We know that is disc is rotating and having angular acceleration then it will have two acceleration .one is radial acceleration and other one is tangential acceleration.
So
[tex]Radial\ acceleration(a_r)=\omega ^2r\ m/s^2[/tex]
[tex]Radial\ acceleration(a_r)=0.6^2 \times0.2 \ m/s^2[/tex]
[tex](a_r)=0.072 \ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=\alpha r\ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=0.3\times 0.2\ m/s^2[/tex]
[tex](a_t)=0.06 \ m/s^2[/tex]
So the total acceleration ,a
[tex]a=\sqrt{a_t^2+a_r^2}\ m/s^2[/tex]
[tex]a=\sqrt{0.06^2+0.072^2}\ m/s^2[/tex]
[tex]a=0.0937 \ m/s^2[/tex]
A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the tangent of the surface of the earth. Assuming the radius of the earth to be 6378 km and that is 5.976 * 10^6 kg, determine the eccentricity of the orbit.
Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore[tex] \theta angle = 0[/tex]
and radial component of given velocity is zero
we have[tex] h = r_o v_r_o = 6378+600 =6.97*10^6 m[/tex]
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
[tex]\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)[/tex]
[tex]GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s[/tex]
so
[tex]\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)[/tex]
solvingt for [tex] \epsilon)[/tex]
[tex]\epsilon = 0.22)[/tex]
therefore eccentrcity of orbit is 0.22
A 4-kg-plastic tank that has a volume of 0.2 m^3 is filled with liquid water. Assuming the density of water is 1000 kg/m^3, determine the weight the combined system.
Answer:
The weight of the combined system is 2001.24 Newtons
Explanation:
From the basic relation between mass, density and volume we know that
[tex]density=\frac{Mass}{Volume}[/tex]
In our context we are given that the density of water is 1000 kg per cubic meters
Thus we can find the mass of 0.2 cubic meters of water using the above relation as
[tex]1000kg/m^{3}=\frac{Mass}{0.2m^{3}}\\\\\therefore Mass=1000kg/m^{3}\times 0.2m^{3}=200kg[/tex]
Hence the mass of water in the tank is 200 kilograms.
The total mass of water and the plastic tank thus becomes
[tex]200+4=204kg[/tex]
Now we know that weight of any given mass is calculated as
[tex]Weight=mass\g[/tex]
where,
'g' is the acceleration due to gravity with value = [tex]9.81m/s^{2}[/tex]
Applying the values in the above equation we get
[tex]Weight=204\times 9.81=2001.24Newtons[/tex]
At the beginning of the compression process of an air-standard Diesel cycle, p1 = 95 kPa and T1 = 300 K. The maximum temperature is 2100 K and the mass of air is 12 g. For a compression ratio of 18, determine the net work developed in kJ (enter a number only)
Answer:
6.8 kJ
Explanation:
p1 = 95 kPa
T1 = 300 K
T3 = 2100 K
m = 12 g
Ideal gas equation:
p * v = R * T
v = R * T / p
R for air is 0.287 kJ/(kg K)
v1 = 0.287 * 300 / 95 = 0.9 m^3/kg
v2 = v1 / cr
v2 = 0.9 / 18 = 0.05 m^3/kg
Assuming an adiabatic compression
p*v^k = constant
k is 1.4 for air
p1 * v1 ^ k = p2 * v2 ^ k
p2 = p1 * (v1 / v2) ^k
p2 = p1 * cr^k
p2 = 95 * 18^1.4 = 5.43 MPa
p1*v1/T1 = p2*v2/T2
T2 = p2*v2*T1/(p1*v1)
T2 = 5430 * 0.05 * 300 / (95 * 0.9) = 952 K
The first principle of thermodynamics
Q = W + ΔU
Since this is an adiabatic process Q = 0
W = -ΔU
W1-2 = -m * Cv * (T2 - T1)
The Cv of air is 0.72 kJ/kg
W1-2 = -0.012 * 0.72 * (952 - 300) = -5.63 kJ
Next the combustion happens and temperature increases suddenly.
v3 = v2 = 0.05 m^3/kg
T2 * p2^((1-k)/k) = T3 * p3^((1-k)/k)
p3 = p2 * (T2/T3)^(k/(1-k)
p3 = 5430 * (952/2100)^(1.4/(1-1.4) = 86.5 MPa
The work is zero because the piston doesn't move.
Next it expands adiabatically:
v4 = v1 = 0.9 m^3/kg
T * v^(k-1) = constant (adiabatic process)
T3 * v3^(k-1) = T4 * v4^(k-1)
T4 = T3 * (v3 / v4)^(k-1)
T4 = 2100 * (0.05 / 0.9)^(1.4-1) = 661 K
p3*v3/T3 = p4*v4/T4
p4 = p3*v3*T4/(v4*T3)
p4 = 86500*0.05*661/(0.9*2100) = 1512 kPa
L3-4 = -m * Cv * (T4 - T3)
L3-4 = -0.012 * 0.72 * (661 - 2100) = 12.43 kJ
Net work:
L1-2 + L3-4 = -5.63 + 12.43 = 6.8 kJ
The Air Force One (Boeing 747-200) has a long-range mission takeoff gross load 833,000 pounds (Ibm). What is Air Force One's takeoff mass in: a. gram (9) b. kilogram (kg) c. tonne (ton) d. Mton
Answer:
The mass in:
1) Grams =[tex]377765.5\times 10^{3}grams[/tex]
2) Kilograms =[tex]377765.5kilograms[/tex]
3)Tonnes =[tex]377.7655tonnes[/tex]
4) Megatonnes =[tex]0.378megatonnes[/tex]
Explanation: The given mass of the aircraft in pounds is 833,000 pounds.
Part 1)
Since we know that 1 pound equals 453.5 grams thus by ratio we have
833,000 pounds =[tex]833000lb\times 453.5\frac{g}{lb}=377765.5\times 10^{3}grams[/tex]
Part 2)
Since we know that 1000 grams equals 1 kilogram
thus the above mass in kilograms equals[tex]\frac{377765.5\times 10^{3}}{1000}=377765.5kg[/tex]
Part 3)
Since there are 1000 kilograms in 1 tonne
thus the given mass is converted into tonnes as
[tex]mass_{tonnes}=\frac{377765.5kg}{1000}=377.765tonnes[/tex]
Part 4)
Since 1 Mega tonne(Mton) equals 1000 tonnes thus the given mass is converted into mega tonnes as
[tex]mass_{M\cdot tonn}=\frac{377.7655tonnes}{1000}=0.378Megatonnes[/tex]
A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.
Answer:
1) Angle with x-axis = 42.03 degrees
2) Angle with y-axis =68.2 degrees
3) Angle with z-axis = 56.14 degrees
Explanation:
given any vector [tex]\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}[/tex]
and any x axis the angle between them is given by
[tex]\theta_x =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{i}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{x\cdot i}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_x=cos^{-1}(\frac{160}{\sqrt{160^2+80^2+120^2}} )=42.03^{o}[/tex]
Angle between the vector and y axis is given by
[tex]\theta_y =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{j}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_y=cos^{-1}(\frac{y\cdot j}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_x=cos^{-1}(\frac{80}{\sqrt{160^2+80^2+120^2}} )=68.2^{o}[/tex]
Similarly angle between z axis and the vector is given by
[tex]\theta_z =cos^{-1}(\frac{\overrightarrow{r}\cdot\widehat{k}}{\sqrt{x^2+y^2+z^2}} )\\\\\theta_x=cos^{-1}(\frac{z\cdot k}{\sqrt{x^2+y^2+z^2}} )[/tex]
Applying values we get
[tex]\theta_z=cos^{-1}(\frac{120}{\sqrt{160^2+80^2+120^2}} )=56.145^{o}[/tex]
The angles made by F will be "42.03°", "68.2°" and "56.14°".
Force and Vector:According to the question,
Force, F = 160 i + 80 j + 120 kN
Let any vector,
[tex]\vec r = x \hat i + y \hat j+ z \hat k[/tex]
The angle between x-axis be:
[tex]\Theta_x[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat i}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{x.i}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{160}{\sqrt{160^2+80^2+120^2} }[/tex])
= 42.03°
and,
The angle between y-axis be:
[tex]\Theta_y[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat j}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{y.i}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{80}{\sqrt{160^2+80^2+120^2} }[/tex])
= 68.2°
and,
The angle between z-axis be:
[tex]\Theta_z[/tex] = Cos⁻¹ ([tex]\frac{\vec r. \hat k}{\sqrt{x^2 +y^2 + z^2} }[/tex])
= Cos⁻¹ ([tex]\frac{z.k}{\sqrt{x^2+y^2+z^2} }[/tex])
By substituting the values,
= Cos⁻¹ ([tex]\frac{120}{\sqrt{160^2+80^2+120^2} }[/tex])
= 56.145°
Thus the above approach is correct.
Find out more information about force here:
https://brainly.com/question/1421935
A closed, rigid tank contains 2 kg of water initially at 80 degree C and a quality of 0.6. Heat transfer occurs until the tank contains only saturated vapor at a higher pressure. Kinetic and potential energy effects are negligible. For the water as the system, determine the amount of energy transfer by heat, in kJ.
Answer:
[tex]Q=1752.3kJ[/tex]
Explanation:
Hello,
In this case, the transferred heat is defined via the first law of thermodynamics as shown below as it is about a rigid tank which does not perform any work:
[tex]Q_{in}=m_{H_2O}(u_2-u_1)[/tex]
The internal energy at the first state (80°C as a vapor-liquid mixture) is computed based on its quality as follows:
[tex]u_1=334.97kJ/kg+0.6*2146.6kJ/kg=1622.93kJ/kg[/tex]
Now, the specific volume turn out into:
[tex]v_1=0.001029m^3/kg+0.6*3.404271m^3/kg=2.0435916m^3/kg[/tex]
As the volume does not change due to the fact that this is about a rigid tank, we must look for a temperature at which the saturated vapor's volume matches with the previously computed volume. This turn out into a temperature of about 94.17 °C at which the internal energy of the saturated vapor is about (by interpolation):
[tex]u_2=2499.1kJ/kg[/tex]
In such a way, the energy transfer by heat is:
[tex]Q=2kg*(2499.1kJ/kg-1622.93kJ/kg)\\Q=1752.3kJ[/tex]
Best regards.
Yield strength (Sy) is typically defined as the point on the stress–strain curve with a strain of during a tension test. a) 0.1% b) 0.2% c) 0.3% c) 0.4%
Answer:
The correct answer is option 'b': 0.2%
Explanation:
The yield strength of a material is defined as the stress in the material when the material begin's to undergo plastic deformation which is also known as yielding.
For materials with well defined yield point such as steel of grade Fe-250 the yield strength can be properly identified from the stress strain curve. But for high strength steel such as Fe-415, Fe-500 the yield point is not properly identified hence the yield strength is taken at 0.2% of proof strain.
The flow curve for a certain metal has parameters: strain-hardening
exponent is 0.22 and strength coefficient is54,000
lb/in2 . Determine:
a) the true stress at a true strain = 0.45
b) the true strain at a true stress = 40,000
lb/in2.
Answer:(a)[tex]45,300.24 lb/in/^2[/tex]
(b)0.255
Explanation:
Given
Strain hardening exponent(n)=0.22
Strength coefficient(k)[tex]=54000 lb/in/^2[/tex]
and we know
[tex]\sigma =k\left ( \epsilon \right )^n[/tex]
where
[tex]sigma =true\ stress[/tex]
[tex]\epsilon =true\ strain[/tex]
(a)True strain=0.45
[tex]\sigma =54000\times 0.45^{0.22}[/tex]
[tex]\sigma =45,300.24 lb/in^2[/tex]
(b)true stress[tex]=40,000 lb/in^2[/tex]
[tex]40000=54000\times \epsilon ^{0.22}[/tex]
[tex]\epsilon ^{0.22}=0.7407[/tex]
[tex]\epsilon =0.7407^{4.5454}=0.255[/tex]
What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?
Answer:
Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.
Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.
Mathematically in a plane AB the shearing stresses are given by
[tex]\tau =\frac{Fcos(\theta )}{A}[/tex]
Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.
Consider two closed systems A and B. System A contains 300 kJ of thermal energy at 20C, whereas system B contains 200kJ of thermal energy at 50oC. Now the systems are brought into contact with each other. Determine the direction of any heat transfer between the two systems and explain your answer.
Answer:
Explanation:
Heat will flow from system B to system A. This is because system B has a higher temperature than system A. Temperature is a measurement od thermodynamic equilibrium. A difference of temperature between two systems is a thermal unbalance, which if they are in contact is compensated by a flow of heat to change the temperatures and reach an equilibrium.
A tow truck is using a cable to pull a car up a 15 degree hill. If the car weighs 4000 lbs and the cable has a diameter of .75 inches, find the stress in the cable when the truck comes to a stop while on the hill. Ignore friction between the car and the pavement.
Answer:40.603 MPa
Explanation:
Given
Car weighs (m)4000 lbs [tex]\approx 1814.37 kg[/tex]
diameter of cable(d)[tex]=0.75 in.\approx 19.05 mm[/tex]
Hill angle [tex]\theta =15^{\circ}[/tex]
Now
tension in cable will bear the weight of car acting parallel to rope which is [tex]mgsin\theta [/tex]
Thus
[tex]T=mgsin\theta [/tex]
[tex]T=1814.37\times 9.81\times sin(15)=11,574.45 N[/tex]
T=11.57 kN
thus stress([tex]\sigma [/tex])=[tex]\frac{T}{A}[/tex]
where A=cross section of wire[tex]=285.059 mm^2[/tex]
[tex]\sigma =\frac{11574.45}{285.059}=40.603 MPa[/tex]
Higher molecular weight results in worst mechanical properties. a)-True b)- false?
Answer:
b)False
Explanation:
Higher molecular weight will increase the mechanical properties because if molecular weight is more then it will require more energy break the molecule .It means that mechanical properties is also improve.
Higher molecular weight will also improve the resistance to corrosion and reduce the chances of material from oxidation.
Higher molecular weight will also increase the viscosity of material and reduce the fluidity.
a piston moves a 25kg hammerhead vertically down 1m from rest to a
velocity of 50m/s in a stamping machine.
what is the total change in energy of the hammerhead?
Answer:
Total change in energy = 31 KJ.
Explanation:
Mass m=25 kg
Height h = 1 m
Initial velocity = 0
Final velocity = 50 m/s
Energy at initial condition
[tex]E_1=mgh+\dfrac{1}{2}mv^2[/tex]
[tex]E_1=25\times 10\times 1+0[/tex]
[tex]E_1=250\ J[/tex]
Energy at final condition
[tex]E_2=0+\dfrac{1}{2}\times 25\times 50^2[/tex]
[tex]E_2=31250\ J[/tex]
So the change in energy = 31250 -250 J
The total change in energy = 31000 J
Answer:
Change in energy will be 31.25 KJ
Explanation:
We have given mass of the piston = 25 kg
Initial velocity u = 0 m/sec
Final velocity v = 50 m/sec
Kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex]
We have to find the change in energy
So change in energy = final KE - initial KE
[tex]\Delta E=\frac{1}{2}m(v^2-u^2)[/tex]
[tex]\Delta E=\frac{1}{2}25\times (50^2-0^2)=31250j=31.25KJ[/tex]
So change in energy will be 31.25 KJ
You live on a street that runs East to West. You just had 2 inche of snow and you live on the North side of the street. You return from class at 2PM and notice all the snow on your sidewalk is gone but across the street it is still there. No one removed the snow. How did it go away?
Answer:
The heat from the sun melted it
Explanation:
If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.
What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ*dV/dy? Then, by dimensional analysis, show that the shear stress has the same units as momentum divided by (area*time).What are the unit or dimensions of viscosity?
Answer:
1) Dimensions of shear rate is [tex][T^{-1}][/tex] .
2)Dimensions of shear stress are [tex][ML^{-1}T^{-2}][/tex]
Explanation:
Since the dimensions of velocity 'v' are [tex][LT^{-1}][/tex] and the dimensions of distance 'y' are [tex][L][/tex] , thus the dimensions of [tex]\frac{dv}{dy}[/tex] become
[tex]\frac{[LT^{-1}]}{[L]}=[T^{-1}][/tex] and hence the units become [tex]s^{-1}[/tex].
Now we know that the dimensions of coefficient of dynamic viscosity [tex]\mu [/tex] are [tex][ML^{-1}T^{-1}][/tex] thus the dimensions of shear stress can be obtained from the given formula as
[tex][\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}][/tex]
Now we know that dimensions of momentum are [tex][MLT^{-1}][/tex]
The dimensions of [tex]Area\times time[/tex] are [tex][L^{2}T][/tex]
Thus the dimensions of [tex]\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}][/tex]
Which is same as that of shear stress. Hence proved.
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C, what will be the percentage change in its volume? If the vat has a diameter of 2 m, how much will the water level rise due to this temperature increase?
Answer:
percentage change in volume is 2.60%
water level rise is 4.138 mm
Explanation:
given data
volume of water V = 500 L
temperature T1 = 20°C
temperature T2 = 80°C
vat diameter = 2 m
to find out
percentage change in volume and how much water level rise
solution
we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure
and we know that is also in term of change in density also
so
E = [tex]-\frac{dp}{dV/V}[/tex] ................1
And [tex]-\frac{dV}{V} = \frac{d\rho}{\rho}[/tex] ............2
here ρ is density
and we know ρ for 20°C = 998 kg/m³
and ρ for 80°C = 972 kg/m³
so from equation 2 put all value
[tex]-\frac{dV}{V} = \frac{d\rho}{\rho}[/tex]
[tex]-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}[/tex]
dV = 0.0130 m³
so now % change in volume will be
dV % = [tex]-\frac{dV}{V}[/tex] × 100
dV % = [tex]-\frac{0.0130}{500*10^{-3} }[/tex] × 100
dV % = 2.60 %
so percentage change in volume is 2.60%
and
initial volume v1 = [tex]\frac{\pi }{4} *d^2*l(i)[/tex] ................3
final volume v2 = [tex]\frac{\pi }{4} *d^2*l(f)[/tex] ................4
now from equation 3 and 4 , subtract v1 by v2
v2 - v1 = [tex]\frac{\pi }{4} *d^2*(l(f)-l(i))[/tex]
dV = [tex]\frac{\pi }{4} *d^2*dl[/tex]
put here all value
0.0130 = [tex]\frac{\pi }{4} *2^2*dl[/tex]
dl = 0.004138 m
so water level rise is 4.138 mm
Derive the dimensions of specific heat that is defined as the amount of heat required to elevate the temperature of an object of mass 1 kg by 1 degree Celcius.
Answer:
Dimension of specific heat will be [tex]=L^2T^{-2}\Theta ^{-1}[/tex]
Explanation:
We know that heat [tex]Q=mc\Delta T[/tex], Q is heat generated, m is mass, c is specific heat and [tex]\Delta T[/tex] is temperature difference
From formula we can write [tex]c=\frac{Q}{m\times \Delta T}[/tex]
Now unit of Q is joule or N-m
Newton can be written as [tex]kgm/sec^2[/tex]
So unit of Q is [tex]kgm^2/sec^2[/tex]
For dimension we use M for kg, L for meter(m) ,T for sec and [tex]\Theta[/tex] for temperature
So dimension of Q is [tex]ML^2T^{-2}[/tex]
So dimension of specific heat will be [tex]\frac{ML^2T^{-2}}{M\Theta }=L^2T^{-2}\Theta ^{-1}[/tex]
Define drag and lift forces.
Explanation:
Drag is the force which is generated parallel and also in opposition to direction of the travel for the object which is moving through the fluid.
Lift is the force which is generated perpendicular to direction of the travel for the object moving through the fluid.
Both of the two forces which are the lift and the drag force act through center of the pressure of object.
The charpy test determines?
Answer:
The charpy test is used to determine amount of energy a material absorbs at fracture.
Explanation:
Charpy Impact test was developed by a French scientist to determine the amount of energy a material absorbs at fracture hence giving the toughness of the material. It is widely used in industrial applications since it is easy to perform and does not requires sophisticated equipment to perform.
The test is performed when a swinging pendulum of known weight is dropped from a known height and is made to strike the metal specimen which is notched.The notch in the sample affects the results of the test hence the notch should be standardized while performing the test. The qualitative results obtained can also be used to compare ductility of different materials.
If a plus sight of 12.03 ft is taken on BM A, elevation 312.547 ft, and a minus sight of 5.43 ft is read on point X, calculate the HI and the elevation of point X.
Answer:
Therefore, height of instrument is 324.577 ft
Therefore, elevation of point x is 330 m
Explanation:
Given that
Plus sight on BM = 12.03 ft
Minus sight is = 5.43 ft
Elevation = 312.547 ft
Height of instrument is H.I
H.I = elevation on bench mark + plus sight
= 312.547 + 12.03 = 324.577 ft
Therefore, height of instrument is 324.577 ft
Elevation at point x is = H.I - minus sight
= 324.577 - (- 5.43)
= 330.00 m
Therefore, elevation of point x is 330 m
The pulley has mass 12.0 kg, outer radius Ro=250 mm, inner radius Ri=200 mm, and radius of gyration kO=231 mm. Cylinder A weighs 71.0 N. Assume there is no friction between the pulley and its axle and that the rope is massless. At the instant when ω=69.0 rad/s clockwise, what is the kinetic energy of the system?
The total kinetic energy of the system is approximately 17128.26 J.T
What is the energy?
Calculate the moment of inertia (I) of the pulley:
= 0.5 * mass * (outer radius² + inner radius²)
= [tex]0.5 * 12.0 kg * ((0.250 m)² + (0.200 m)²)[/tex]
= 1.925 kg * m²
Use the parallel-axis theorem to find I:
I = [tex]1.925 kg * m² + 12.0 kg * (0.231 m)²[/tex]
I = 2.5683 kg * m²
Calculate the kinetic energy of the pulley:
= 0.5 * I * omega²
= [tex]0.5 * 2.5683 kg * m² * (69.0 rad/s)[/tex]
= 6555.63 J
Calculate the linear velocity of cylinder A:
v = outer radius * omega
v =[tex]0.250 m * 69.0 rad/s[/tex]
v = 17.25 m/s
Calculate the kinetic energy of cylinder A:
= 0.5 * mass * v²
= [tex]0.5 * 71.0 kg * (17.25 m/s)²[/tex]
= 10572.63 J
KEtotal = KEpulley + KEcylinder
= [tex]6555.63 J + 10572.63 J[/tex]
= 17128.26 J
a) What is the Damkohler number? b) What is the significance of a system with a low Damkohler number?
Explanation:
Damkohler numbers are mainly used in chemistry. It is a dimensionless number. It denotes the timescale at which the reaction takes place with relation to the transport phenomenon.
There are two Damkohler numbers
First Damkohler number is the ratio of reaction rate to the convective mass transport rate.
[tex]Da=\frac{\text{Reaction rate or chemical reaction timescale}}{\text{Convective mass transport rate}}[/tex]
Second Damkohler number is the ratio of reaction rate to the diffusive mass transfer rate
[tex]Da_{II}=\frac{\text{Reaction rate or chemical reaction timescale}}{\text{Diffusive mass transfer rate}}[/tex]
It can be seen from the equations that if the numerator is greater than the denominator then Da>1 and vice versa.
So,
When Da>1, the diffusion rate distribution is lower than the reaction rate.
When Da<1, the reaction rate is lower than the diffusion rate.
The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.
Answer:
Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]
Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]
[tex]g_{s} = 8.79076[/tex]
[tex]w_{w} = 9782.36 N/m^{3}[/tex]
Given:
Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]
Mass of a can of pop , m = 0.369 kg
Weight of an empty can, W = 0.153 N
Solution:
Now, weight of a full can pop, W
W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]
Now weight of the pop in can is given by:
w = W' - W = 3.616 - 0.513 = 3.103 N
Now,
The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]
[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]
Now, density of the pop:
[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]
[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]
Now,
Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]
where
[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]
Now, for water at [tex]20^{\circ}c[/tex]:
Specific density of water = [tex]998.2 kg/m^{3}[/tex]
Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]
Specific weight of water at [tex]20^{\circ}c[/tex]:
[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]
A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates with a maximum amplitude of 10 mm. What is its natural frequency?
Answer:
f=1.59 Hz
Explanation:
Given that
Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.
Velocity = 100 mm/s
Maximum amplitude = 10 mm
We know that for a simple undamped system spring mass system
[tex]V_{max}=\omega A[/tex]
now by putting the values
[tex]V_{max}=\omega A[/tex]
100 = ω x 10
ω = 10 rad/s
We also know that
ω=2π f
10 = 2 x π x f
f=1.59 Hz
So the natural frequency will be f=1.59 Hz.
Give examples of engineering structures which can be modelled as thin walled cylinders.
Answer:
Pipes, pressure vessels, tanks, reactors, tubes and nozzles
Explanation:
Thin walled cylinders are typically defines as having wall thickness of 1/10 of the radius (doesn't matter much if inner or outer, they should be similar). Also this is used mostly for things that will be subject to some radial load, as opposed to axles and shafts.
As such, some structures that can be modeled as thin walled cylinders are pipes, pressure vessels, tanks, reactors, tubes and nozzles.
A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature of the steam at the turbine inlet is 400 oC, and the mass flow rate of steam through the cycle is 37 kg/s. Determine: a) the thermal efficiency of the cycle (%) and b) the net power output of the power plant (kW).
Answer:
a)31%
b)34MW
Explanation:
A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.
This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image
To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.
• The pressure of state 1 and 4 are equal
• The pressure of state 2 and 3 are equal
• State 1 is superheated steam
• State 2 is in saturation state
• State 3 is saturated liquid at the lowest pressure
• State 4 is equal to state 3 because the work of the pump is negligible.
Once all enthalpies are found, the following equations are used using the first law of thermodynamics
Wout = m (h1-h2)
Qin = m (h1-h4)
Win = m (h4-h3)
Qout = m (h2-h1)
The efficiency is calculated as the power obtained on the heat that enters
Efficiency = Wout / Qin
Efficiency = (h1-h2) / (h1-h4)
For this problem, we will first find the enthalpies in all states
h1=3231kJ/Kg
h2=2310kJ/Kg
h3=h4=272kJ/Kg
A) using the eficiency ecuation
Efficiency = (h1-h2) / (h1-h4)
Efficiency =(3231-2310)/(3231-272)=0.31=31%
b)using ecuation for Wout
Wout = m (h1-h2)
Wout=37(3231-2310)=34077KW=34.077MW
1.19. A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m·s−2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?\
Answer:
a) 19094 N
b) 110.055 kPa
c) 1222 J
Explanation:
The force on the gas is the weight plus the atmospheric pressure multiplied by the piston area
F = P + p * A
F = m * g + p * π/4 * d^2
F = 150 * 9.813 + 101570 * π/4 * 0.47^2 = 19094 N
The pressure is the force divided by the area of the piston
p1 = F / A
p1 = F / (π/4 * d^2)
p1 = 19094 / (π/4 * 0.47^2) = 110055 Pa = 110.055 kPa
variation of gravitational potential energy is defined as
ΔEp = m * g * Δh
ΔEp = 150 * 9.813 * 0.83 = 1222 J
In this exercise we have to use the knowledge of force to calculate the required energies, so we have to:
a) 19094 N
b) 110.055 kPa
c) 1222 J
What is the concept of force?In the field of physics, force is a physical action that causes deformation or that changes the state of rest or movement of a given object.
a) Knowing that the force formula is defined by:
[tex]F = P + p * A\\F = m * g + p *\pi /4 * d^2\\F = 150 * 9.813 + 101570 * \pi /4 * 0.47^2 = 19094 N[/tex]
b) Knowing that the force exerted by an area is equal to the pressure in that area, we have:
[tex]p_1 = F / A\\p_1 = F / (\pi /4 * d^2)\\p_1 = 19094 / (\pi /4 * 0.47^2) = 110055 Pa = 110.055 kPa[/tex]
c)So calculating the potential energy we have:
[tex]\Delta E_p = m * g * \Delta h\\\Delta E_p = 150 * 9.813 * 0.83 = 1222 J[/tex]
See more about force at brainly.com/question/26115859
The higher the degree of polymerization, the longer the chains will be, which results in a lower chain molecular weight. a)-True b)- false?
Answer:
b)False
Explanation:
When one or more than one monomer join to other monomer then they form a chain and this joining of monomers is called degree of polymerization .
Degree of polymerization :
[tex]Degree\ of\ polymerization=\dfrac{Mass\ of\ polymer}{Mass\ of\ monomer}[/tex]
So from above we can say that when chain will become longer then the weight of polymer will increase.
A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a) 0.246m (b) 0.318m (c) 1.3m (d) 0.052m (e) None of the above
Answer:
The correct answer is option 'a': 0.046 meters.
Explanation:
We know that the exit velocity of a jet of water is given by Torricelli's law as
[tex]v=\sqrt{2gh}[/tex]
where
'v' is velocity of head
'g' is acceleration due to gravity
'h' is the head under which the jet falls
Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.
Applying the values in above equation we get the exit velocity as
[tex]v=\sqrt{2\times 9.81\times 90}=42.02m/s[/tex]
now we know the relation between discharge and velocity as dictated by contuinity equation is
[tex]Q=V\times Area[/tex]
Applying values in the above equation and solving for area we get
[tex]Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}[/tex]
The circular area is related to diameter as
[tex]Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m[/tex]
Thus the diameter of the nozzle is 0.246 meters
A certain working substance receives 100 Btu reversibly as heat at a temperature of 1000℉ from an energy source at 3600°R. Referred to as receiver temperature of 80℉, calculate: A. the available energy of the working substance ( 63 Btu ) B. the available portion of the 100 Btu added at the source temperature ( 85 Btu ) C. the reduction in available energy between the source temperature and the 1000℉ temperature ( 22 Btu )
Answer:
Explanation:
t1 = 1000 F = 1460 R
t0 = 80 F = 540 R
T2 = 3600 R
The working substance has an available energy in reference to the 80F source of:
B1 = Q1 * (1 - T0 / T1)
B1 = 100 * (1 - 540 / 1460) = 63 BTU
The available energy of the heat from the heat wource at 3600 R is
B2 = Q1 * (1 - T0 / T2)
B2 = 100 * (1 - 540 / 3600) = 85 BTU
The reduction of available energy between the source and the 1460 R temperature is:
B3 = B2 - B1 = 85 - 63 = 22 BTU