Answer:
Equilibrium of a system is defined as when the concentration of reactants and products of a chemical reaction are in equilibrium and their ratio does not vary. In another words we can say that, when the forward reaction rate is similar to the reverse reaction rate.
Thermodynamics equilibrium is defined as in an isolated system when there is no change in energy and entropy. It is determined by the its intensive properties like volume and pressure.
An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R. Are these measurements reasonable? Why?
The maximum efficiency must be greater than the actual efficiency of the heat engine. These measurements are not reasonable.
What is Efficiency?The efficiency is defined as the work done by the engine divided by the heat supplied.
So, maximum efficiency η = 1 - T₁/T₂ = W/Qs
Where T₁ is the lower temperature and T₂ is the higher temperature.
An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R
Substitute the value into the expression , we get
η = 1 -(540 / 900) x 100 %
η = 40 %
η = Work done/heat supplied
Substitute the value into the expression , we get
η = 160/300 x 100 %
η = 53 %
So, the maximum efficiency must be greater than the actual efficiency of the heat engine.
Thus, these measurements are not reasonable.
Learn more about efficiency.
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The experimentalist's claim that the heat engine converts more work than allowed by the Carnot efficiency is not reasonable because it exceeds the theoretical maximum efficiency, violating the second law of thermodynamics.
Explanation:The experimentalist's claim that a heat engine receives 300 Btu of heat from a source at 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R can be assessed using the second law of thermodynamics and the concept of Carnot efficiency. The Carnot efficiency can be calculated using the temperatures of the hot and cold reservoirs (Th and Tc):
Carnot Efficiency (EffC) = 1 - (Tc/Th)
Assuming the temperatures are in degrees Rankine (where 0 R is absolute zero), we can calculate the maximum theoretical efficiency for a perfectly reversible engine:
EffC = 1 - (540 R / 900 R) = 1 - 0.6 = 0.4 (or 40%)
However, the experimentalist's claim states that 160 Btu is converted to work out of the 300 Btu received, which represents an actual efficiency (EffA) of:
EffA = Work done / Heat received = 160 Btu / 300 Btu = 0.5333 (or 53.33%)
Since the actual efficiency (53.33%) claimed by the experimentalist exceeds the maximum possible Carnot efficiency (40%) for a heat engine operating between these temperatures, the experimentalist's measurements are not reasonable and violate the second law of thermodynamics.
Prove that the slope of a constant-volume line is steeper than that for a constant-pressure line for a given state (point) on the T-s diagram.
Explanation:
We know that first T-ds equation
Tds=[tex]C_v[/tex]dT+Pdv
For constant volume process dv=0
⇒Tds=[tex]C_v[/tex]dT
So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_v}[/tex] ----(1)
We know that second T-ds equation
Tds=[tex]C_p[/tex]dT-vdP
For constant pressure process dP=0
⇒Tds=[tex]C_p[/tex]dT
So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_p}[/tex] ---(2)
We know that [tex]C_p[/tex] is always greater than [tex]C_v[/tex].
So from equation (1) and (2) we can say that slope of constant volume process will be always greater than constant pressure process on T-s diagram.
The two stroke engines has greater cooling and lubrication requirements than four stroke engine. Explain why?
Answer:
A two stroke engine produces twice the power compared to a four stroke engine of same weight and size.
Explanation:
In a two stroke engine, all the four processes namely, intake stroke, compression stroke, power stroke and exhaust stroke takes place in one revolution of crankshaft or two strokes of the piston. While in a four stroke engine, all the four processes namely intake stroke, compression stroke, power stroke and exhaust stroke take place in two revolution of crankshaft or four strokes of the piston.
Therefore, there is one power stroke in one revolutions of the crankshaft in case of a two stroke engine as compared to the four stoke engine where there is one power stroke for two revolutions of the crank shaft.
So the power developed in a two stroke engine is more ( nearly twice ) as compared to a four stroke engine of the same capacity. When power produced is more, the heat dissipation is also more in case of a two stroke engine. So greater cooling is required to dissipate heat from a two stroke engine as compared to a four stroke engine.
Also in a two stroke engine, the lubricating oil is used with the oil whereas a four stroke engine has a separate tank for lubricating oil. So the lubricating oil gets burnt quickly in a two stroke engine.
Thus, to dissipate more heat, a two stroke engines has greater cooling and lubrication requirements than a four stroke engines as power produce in a two stroke engine is more than a four stoke engine with same weight or size.
What is the difference between the drag coefficient and skin friction coefficient for flow along flat plate.
Answer:
Drag Coefficient:
It is a dimensionless quantity that is used to quantify air friction or fluid friction i.e., drag and is commonly denoted by [tex]C_{d}[/ tex].It is used in Drag Equation, where a lower drag coefficient indicates that the object will have lower value of air friction or fluid friction(i.e., aerodynamic or hydrodynamic drag). It is always associated with a particular surface area.
Drag Coefficient is given by:
[tex]C_{d} = \frac{2D}{A\rho v^{2}}[/tex]
Skin Friction Coefficient:
It is a dimensionless quantity which represents skin shear stress due to dynamic pressure of a free stream.
Skin Friction Coefficient is given by:
[tex]C_{f} = \frac{2\tau _{w}}{\rho V^{2}}[/tex]
where,
[tex]\tau _{w}[/tex]= skin shear stress
v = free stream speed
Define Viscosity. What are the main differences between viscous and inviscid flows?
1. Define Viscosity
In physics, Viscosity refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.
What are the main differences between viscous and inviscid flows?
Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density
Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch diameter pipe. What is the head loss through this contraction
Answer:
Head loss=0.00366 ft
Explanation:
Given :Water flow rate Q=0.15 [tex]\frac{ft^{3}}{sec}[/tex]
[tex]D_{1}[/tex]= 6 inch=0.5 ft
[tex]D_{2}[/tex]=2 inch=0.1667 ft
As we know that Q=AV
[tex]A_{1}\times V_{1}=A_{2}\times V_{2}[/tex]
So [tex]V_{2}=\frac{Q}{A_2}[/tex]
[tex]V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}[/tex]
[tex]V_{2[/tex]=0.687 ft/sec
We know that Head loss due to sudden contraction
[tex]h_{l}=K\frac{V_{2}^2}{2g}[/tex]
If nothing is given then take K=0.5
So head loss[tex]h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}[/tex]
=0.00366 ft
So head loss=0.00366 ft
What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=[tex]\frac{s}{(s+3)^2}[/tex]
U(s)=[tex]\frac{1}{s}[/tex] for unit step function
output =H(s)×U(s)
=[tex]\frac{s}{(s+3)^2}[/tex]×[tex]\frac{1}{s}[/tex]
=[tex]\frac{1}{(s+3)^2}[/tex]
taking inverse laplace of output
output=t×[tex]e^{-3t}[/tex]
at t=0 putting the value of t=0 in output
output =0
Modified Reynolds Analogy is often used in the Heat Exchanger industry. a) True b) False
Answer: True
Explanation: Modified Reynolds Analogy is based on heat transfer process in the Plate Heat Exachanger's(PHE) which is done in the phases that have complex form plates shape by being changed into sheet metal. This analogy helps in the increment of the reliability and operation to be carried out easily in the industries.Thus the given statement is true.
Nanocomposite coatings: a)-Are mostly composed of metals b)-They comprise at least two immiscible phases c)-They are used to improve appearance d)-They are used to improve finish
Answer: b) They comprise at least two immiscible phases
Explanation: Nano composite coating are the coating that are the mixture of one or more phases that are irregular or immiscible in nature. They are in the nano- material form which helps in the improvement of chemical as well as physical properties. The phase that are usually present in nano composite coating are of nano crystalline phase or amorphous phase or two nano crystalline phases which are different from each other.
What is an isentropic process?
Answer: Isentropic process is the process in fluids which have a constant entropy.
Explanation: The isentropic process is considered as the ideal thermodynamical process and has both adiabatic as well as reversible processes in internal form.This process supports no transfer of heat and no transformation of matter .The entropy of the provided mass also remains unchanged or consistent.These processes are usually carried out on material on the efficient device.
A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm and the gap above the bottom plate is 30 mm. The gap between the fixed plates contain SAE 30 oil at 15.6°C. Determine the force needed to push the middle plate at a velocity of 1.0 m/s.
Answer:0.3166N
Explanation:
Given data
Area [tex]\left ( A\right )=500 cm^2[/tex]
Gap below top plate[tex]\left ( y_1\right )=20 mm[/tex]
Gap above bottom plate[tex]\left ( y_2\right )=30 mm[/tex]
SAE 30 oil viscosity =[tex]0.38 N-s/m^2[/tex]
Velocity of middle plate[tex]\left ( v\right )=1 m/s[/tex]
There will viscous force on middle plate i.e. at above surface and below surface
Viscous force[tex]\left ( F\right )=\mu \frac{Av}{y}[/tex]
[tex]Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ][/tex]
[tex]F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ][/tex]
[tex]F=31.66\times 10^{-2}=0.3166 N[/tex]
this is force by oil on plate thus we need to apply atleast 0.3166N to move plate
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T that may be applied to the shaft if the allowable shear stress is equal to 100 MPa.
Answer:
Explanation:
Using equation of pure torsion
[tex]\frac{T}{I_{polar} }=\frac{t}{r}[/tex]
where
T is the applied Torque
[tex]I_{polar}[/tex] is polar moment of inertia of the shaft
t is the shear stress at a distance r from the center
r is distance from center
For a shaft with
[tex]D_{0} =[/tex] Outer Diameter
[tex]D_{i} =[/tex] Inner Diameter
[tex]I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}[/tex]
Applying values in the above equation we get
[tex]I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\
I_{polar}= 1.74[/tex] x [tex]10^{-7} m^{4}[/tex]
Thus from the equation of torsion we get
[tex]T=\frac{I_{polar} t}{r}[/tex]
Applying values we get
[tex]T=\frac{1.74X10^{-7}X100X10^{6} }{.021}[/tex]
T =829.97Nm
As the car gets older, will its engine's compression ratio change, if yes how.
Answer:
yes
Explanation:
yes, it is correct as car gets older the engine's compression ratio changes .
compression ratio is the volume in maximum compression chamber to the volume in the piston at full compression .
More and more engine will be used if the ring wears compression goes down
or there can be one condition if carbon build up on the piston then this can alter the compression ratio.
A block of mass 0.75 kg is suspended from a spring having a stiffness of 150 N/m. The block is displaced downwards from its equilibrium position through a distance of 3 cm with an upward velocity of 2 cm/sec. Determine: a)- The natural frequency b)-The period of oscillation c)-The maximum velocity d)-The phase angle e)-The maximum acceleration
Answer:
a)f=2.25 Hz
b)Time period T=.144 s
c)tex]V_{max}[/tex]=0.42 m/s
d)Phase angle Ф=87.3°
e) [tex]a_{max}=6.0041 [tex]\frac{m}{s^2}[/tex]
Explanation:
a)
Natural frequency
[tex]\omega _n=\sqrt {\dfrac{K}{m}}[/tex]
[tex]\omega _n=\sqrt {\dfrac{150}{0.75}}[/tex]
[tex]\omega _n[/tex]=14.14 rad/s
w=2πf
⇒f=2.25 Hz
b) Time period
[tex]=\dfrac{2π}{\omega _n}[/tex]
T=[tex]\frac{1}{f}[/tex]
Time period T=.144 s
c)Displacement equation
[tex]x=Acos\omega _nt+Bsin\omega _nt[/tex]
Boundary condition
t=o,x=0.03 m
t=0,v=.02m/s , V=[tex]\frac{dx}{dt}[/tex]
Now by using these above conditions
A=0.03,B=0.0014
x=0.03 cos14.14 t+0.0014 sin14.14 t
⇒x=0.03003sin(14.14t+87.3)
[tex]V_{max}=\omega_n X_{max}[/tex]
[tex]V_{max}=14.14\times 0.03003[/tex]=0.42 m/s
d)
Phase angle Ф=87.3°
e)
Maximum acceleration
[tex]a_{max}=(\omega _n )^2X_{max}[/tex]
[tex]a_{max}=(14.14)^20.03003[/tex]=6.0041 [tex]\frac{m}{s^2}[/tex]
Answer:
A. 2.249 hz
B. 0.45 s
C. 0.424 m/s
D. 66⁰
E. 6 m/s^2
Explanation:
Step 1: identify the given parameters
mass of the block (m)= 0.75kg
stiffness constant (k) = 150N/m
Amplitude (A) = 3cm = 0.03m
upward velocity (v) = 2cm/s
Step 2: calculate the natural frequency (F)by applying relevant formula in S.H.M
[tex]f=\frac{1}{2\pi } \sqrt \frac{k}{m}[/tex]
[tex]f=\frac{1}{2\pi } \sqrt \frac{150}{0.75}[/tex]
f = 2.249 hz
Step 3: calculate the period of the oscillation (T)
[tex]period (T) = \frac{1}{frequency}[/tex]
[tex]T = \frac{1}{2.249} (s)[/tex]
T = 0.45 s
Step 4: calculate the maximum velocity,[tex]V_{max}[/tex]
[tex]V_{max} = A\sqrt{\frac{k}{m} }[/tex]
A is the amplitude of the oscilation
[tex]V_{max} = 0.03\sqrt{\frac{150}{0.75} }[/tex]
[tex]V_{max} = 0.424(\frac{m}{s})[/tex]
Step 5: calculate the phase angle, by applying equation in S.H.M
[tex]X = Acos(\omega{t} +\phi)[/tex]
where X is the displacement; calculated below
Displacement = upward velocity X period of oscillation
[tex]displacement (X) = vt (cm)[/tex]
X = (2cm/s) X (0.45 s)
X = 0.9 cm = 0.009m
where [tex]\omega[/tex] is omega; calculated below
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{150}{0.75} }[/tex]
[tex]\omega= 14.142[/tex]
[tex]\phi = phase angle[/tex]
Applying displacement equation in S.H.M
[tex]X = Acos(\omega{t}+\phi)[/tex]
[tex]0.009 = 0.03cos(14.142 X 0.45+\phi)[/tex]
[tex]cos(6.364+\phi) = \frac{0.009}{0.03}[/tex]
[tex]cos(6.364+\phi) = 0.3[/tex]
[tex](6.364+\phi) = cos^{-1}(0.3)[/tex]
[tex](6.364+\phi)= 72.5⁰[/tex]
[tex]6.364+\phi =72.5⁰[/tex]
[tex]\phi[/tex] =72.5 -6.364
[tex]\phi[/tex] =66.1⁰
Phase angle, [tex]\phi[/tex] ≅66⁰
Step 6: calculate the maximum acceleration, [tex]a_{max}[/tex]
[tex]a_{max} = \omega^{2}A[/tex]
[tex]a_{max}[/tex] = 14.142 X 14.142 X 0.03
[tex]a_{max}[/tex] = 5.999 [tex](\frac{m}{s^{2} })[/tex]
[tex]a_{max}[/tex] ≅ 6 [tex](\frac{m}{s^{2} })[/tex]
The drilling pipe on an oil rig is made from steel pipe which having thickness of 5-mm and an outside diameter of 90-mm. Calculate the maximum shear stress occur in the pipe if the pipe is turning at 650 rev/min while being powered using 12kW motor.
Answer:
[tex]\tau_{max}= 3.28 \ MPa[/tex]
Explanation:
outside diameter = 90 mm
inside diameter = 90- 2× t= 90- 2× 5 = 80mm
where t is thickness of pipe.
power (P) = 12 kW
Revolution (N)= 650 rev/min
we
Power = torque × angular velocity
P= T× ω
ω = [tex]\frac{2 \pi N}{60}[/tex]
[tex]P=T \times \frac{2\pi N}{60}\\12 \times 10^3=T\times \frac{2\pi \times 650}{60}[/tex]
T= 176.3 Nm
for maximum shear stress
[tex]\frac{\tau_{max}}{y_{max}}=\frac{T}{I_p}[/tex]
where ymax is maximum distance from neutral axis.
[tex]y_{max}=\frac{d_0}{2}= \frac{90}{2}[/tex]= 45 mm
[tex]I_p[/tex]= polar moment area
= [tex]\frac{\pi}{32} (d_o^4-d_i^4)=\frac{\pi}{32} (90^4-80^4)[/tex]
= 2,420,008 mm⁴
[tex]\dfrac{\tau_{max}}{45}=\dfrac{176.3 \times 10^3}{2,420,008}[/tex]
[tex]\tau_{max}= 3.28 \ MPa[/tex]
Consider an open loop 1-degree-of-freedom mass-spring damper system. The system has mass 4.2 kg, and spring stiffness of 85.9 N/m, and damping coefficient of 1.3 N.s/m. What is the non-dimensional damping ratio of the system? Use at least 4 significant digits after the decimal point.
Answer:
Damping ratio [tex]\zeta =0.0342[/tex]
Explanation:
Given that
m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m
We need to find damping ratio
We know that critical damping co-efficient
[tex]C_c=2\sqrt {mk}[/tex]
[tex]C_c=2\sqrt {4.2\times 85.9}[/tex]
[tex]C_c=37.98[/tex] N.s/m
Damping ratio([tex]\zeta[/tex]) is the ratio of damping co-efficient to the critical damping co-efficient
So [tex]\zeta =\dfrac{C}{C_c}[/tex]
[tex]\zeta =\dfrac{1.3}{37.98}[/tex]
[tex]\zeta =0.0342[/tex]
So damping ratio [tex]\zeta =0.0342[/tex]
What is the difference between a refrigeration cycle and a heat pump cycle?
Answer:
In refrigeration cycle heat transfer from inside refrigeration
In heat pump cycle heat transfer from environment
Explanation:
heat cycle is mechanical process use for cool the temperature but
In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator
A solid 0.75 in diameter steel shaft transmits 7 hp at 3,200 rpm. Determine the maximum shear stress magnitude produced in the shaft. Hint: Use P=Tω and convert hp to ft-lbf/s. Find τ by using Tc/J. Recall max shear stress will be on the outer most surface.
Answer:[tex]\tau _\left ( max\right )[/tex]=11.468MPa
Explanation:
Given data
[tex]power[/tex][tex]\left ( P\right )[/tex]=7 hp=5220 W
N=3200rpm
[tex]\omega [/tex]=[tex]\frac{2\pi\times N}{60}[/tex]=335.14 rad/s
diameter[tex]\left ( d\right )[/tex]=0.75in=19.05mm
we know
P=[tex]Torque\left ( T\right )\omega [/tex]
5220=[tex]T\times 335.14[/tex]
T=15.57 N-m
And
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Polar\ modulus}[/tex]
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Z_{P}}[/tex]
[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{16\times T}{\pi d^{3}}[/tex]
[tex]\tau _\left ( max\right )[/tex]=11.468MPa
Answer:
Maximum shear stress is 11.47 MPa.
Explanation:
Given:
D=.75 in⇒D=19.05 mm
P=7 hp⇒ P=5219.9 W
N=3200 rpm
We know that
P=T[tex]\times \omega[/tex]
Where T is the torque and [tex]\omega[/tex] is the speed of shaft.
P=[tex]\frac{2\pi N\times T}{60}[/tex]
So 5219.9=[tex]\frac{2\pi \times 3200\times T}{60}[/tex]
T=15.57 N-m
We know that maximum shear stress in shaft
[tex]\tau _{max}=\dfrac{16T}{\pi \times D^3}[/tex]
[tex]\tau _{max}=\dfrac{16\times 15.57}{\pi \times 0.01905^3}[/tex]
[tex]\tau _{max}[/tex]=11.47 MPa
So maximum shear stress is 11.47 MPa.
If a point is positioned 16 inches from the origin on the x-y plane, and it is 12 inches above the x axis, what is its position in terms of the unit vectors i and j?
Answer:
[tex]p=12\hat{i}+10.58\hat{j}[/tex]
Explanation:
distance of point from the origin is 16 inch
x-axis distance =12 inch
[tex]p=\sqrt{x^2+y^2[/tex]
[tex]y=\sqrt{p^2-x^2} \\y=\sqrt{16^2-12^2}\\ y=\sqrt{112}[/tex]
y=10.58 inch
[tex]p=12\hat{i}+10.58\hat{j}[/tex]
The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is linear with depth what is the surface velocity of flow in the channel?
Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
[tex]\dot{m}=\rho AV[/tex]
Putting all the value to get the velocity of the flow
[tex]\frac{\dot{m}}{\rho A} = V[/tex]
[tex]V = \frac{4000}{1000*4*2}[/tex]
V = 0.5 m/s
The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion. a)-True b)-False
Answer:
a)-True
Explanation:
The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion.
This statement is true.
What is the function of the following: 1- Oil rings 2- Flywheel 3- Timing gears
Answer:
Explanation:
The functions of the following are as follows:
1). Oil Rings:
Regulates oil within cylinder walls.Helps to keep cylinder walls lubricatedPrevent Heat transferReduce friction between piston and cylinder2). Fly Wheel:
It is used to store rotational energyIt resist any change in rotational speed due to its moment of inertiaIt helps to control the orientation of the mechanical system3). Timing gears:
It provides synchronization in the rotation of crankshaft and the camshaft so as to provide proper valve opening and closing time during each cylinder's stroke(intake and exhaust).A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, determine the magnitude of the particle's acceleration when t= 2 s.
Answer:
4.83m/[tex]s^{2}[/tex]
Explanation:
For a particle moving in a circular path the resultant acceleration at any point is the vector sum of radial and the tangential acceleration
Radial acceleration is given by [tex]a_{radial}=w^{2}[/tex]r
Applying values we get [tex]a_{radial}=(2t)^{2}[/tex]X0.3m
Thus [tex]a_{radial}=1.2t^{2}[/tex]
At time = 2seconds [tex]a_{radial}= 4.8m/s^{2}[/tex]
The tangential acceleration is given by [tex]a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}[/tex]
[tex]a_{tangential}=\frac{d(2tr)}{dt}[/tex]
[tex]a_{tangential}= 2r[/tex]
[tex]a_{tangential}=0.6m/s^{2}[/tex]
Thus the resultant acceleration is given by
[tex]a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}[/tex]
[tex]a_{res} =\sqrt{4.8^{2}+0.6^{2} } =4.83m/s^{2}[/tex]
Which of the following is not a method of heat transfer? A. Conduction B. Convection C. Injection D. Radiation is desirable.
Answer:
C) Injection
Explanation:
Injection is a molding process, not a heat transfer mechanism.
A fluid should be changed a. when its viscosity changes b. when it becomes contaminated c. at operating temperature d. when its acidity increases e. all of the above
Answer:
e.All of the above
Explanation:
A fluid should be change
a.When its acidity increase
b.When its become contaminated
c.At operating temperature
Generally fluid is used in fluid power.Fluid power means transmission of power by using pressurized fluids.
Fluid power works on Pascal's and Bernoulli law.
From the above option we can say that all option is right for fluid .
Cylinder cushions at the ends of the cylinder speeds the piston up at the end of the stroke. a)- True b) False
The efficiency of a transformer is mainly dependent on: a)- Core losses b)- Copper losses c)- Stray losses d)- Dielectric losses
The efficiency of a transformer is mainly dependent on a)- Core losses
Hope this helps! :)
Pascal's law tells us that, pressure is transmitted undiminished throughout an open container. a)- True b) False
Answer:
False
Explanation:
Pascal's law is not for open container it is for enclosed fluid
Pascal's law (also known as the principle of transmission of fluid pressue)
states that when pressure is applied at any part of an enclosed fluid it is transmitted undiminished throughout the fluid as well as to the walls of the container containing the fluid i.e., change in pressure at any point in an enclosed incompressible fluid is transmitted such that the same change occurs everywhere.
Use Newton's method to determine the angle θ, between 0 and π/2 accurate to six decimal places. for which sin(θ) = 0.1. Show your work until you start computing x1, etc. Then just write down what your calculator gives you.
Answer:
x3=0.100167
Explanation:
Let's find the answer.
Because we are going to find the solution for sin(Ф)=0.1 then:
f(x)=sin(Ф)-0.1 and:
f'(x)=cos(Ф)
Because 0<Ф<π/2 let's start with an initial guess of 0.001 (x0), so:
x1=x0-f(x0)/f'(0)
x1=0.001-(sin(0.001)-0.1)/cos(0.001)
x1= 0.100000
x2=0.100000-(sin(0.100000)-0.1)/cos(0.100000)
x2=0.100167
x3=0.100167
The area under the moment diagram is shear force. a)-True b)-False
Answer:
False
Explanation:
as we know that [tex]V(x)=\frac{dM}{dx}\\ \\=> M(x) = \int\limits^x_o {V(x)} \, dx \\\\[/tex]
=> Area under shear diagram gives the moment at any point but the reverse cannot be established from the same relation