Answer:
(a)[tex]1.308\times 10^{-4}m/sec[/tex]
(b)57.33831 pound/cubic feet
(c)120.1095 hp
Explanation:
We have
(a) 760 miles / hour
We know that [tex]1\ mile\ =0.00062m[/tex]
And 1 hour = 60×60=3600 sec
So [tex]760miles/hour=\frac{760\times 0.00062meter}{3600sec}=1.308\times 10^{-4}m/sec[/tex]
(b) 927 kg/cubic meter to mass/cubic foot
We know that 1 kg = 2.20 pound
So 921 kg = 921×2.20=2026.2 pound
We know that 1 cubic meter = 35.31 cubic feet
So 57.33831 pound / cubic feet
(c) We have to convert to hp
[tex]5.37\times 10^3kj/min=\frac{5.37\times 1000kj}{60sec}=89.5kj/sec[/tex]
We know that 1 kj /sec = 1.341 hp
So 89.5 kj/sec = 89.5××1.341=120.1095 hp
What does stall mean?
Answer:
Explanation:
In aeronautics stall is the separation of the boundary layer from the wings of plane. This causes loss of lift because the otherwise low pressure area above the wing become filled with turbulent whirls.
Stall occurs when the critical angle of attack of the wing is exceeded.
An airplane flies horizontally at 80 m/s. Its propeller delivers 1300 N of thrust (forward force) to overcome aerodynamic drag (backward force). Using dimensional reasoning and unity conversion ratios, calculate the useful power delivered by the propeller in units of kW and horsepower.
Answer:
Power in kW is 104 kW
Power in horsepower is 139.41 hp
Solution:
As per the question:
Velocity of the airplane, [tex]v_{a} = 80 m/s[/tex]
Force exerted by the propeller, [tex]F_{p} = 1300 N[/tex]
Now,
The useful power that the propeller delivered, [tex]P_{p}[/tex]:
[tex]P_{p} = \frac{Energy}{time, t}[/tex]
Here, work done provides the useful energy
Also, Work done is the product of the displacement, 'x' of an object when acted upon by some external force.
Thus
[tex]P_{p} = \frac{F_{p}\times x}{time, t}[/tex]
[tex]P_{p} = \frac{F_{p}\times x}{time, t}[/tex]
[tex]P_{p} = F_{p}\times v_{a}[/tex]
Now, putting given values in it:
[tex]P_{p} = 1300\times 80 = 104000 W = 104 kW[/tex]
In horsepower:
1 hp = 746 W
Thus
[tex]P_{p} = \frac{104000}{746} = 139.41 hp[/tex]
A bar of uniform cross section is 86.9 in longand weighs 89.1 lb. A weight of 79.0 lb is suspended from one end. The bar and weight combination is to be suspended from a cable attached at the balance point. How far (in) from the weight should the cable be attached, and what is the tension (lb) in the cable?
Answer:
y = 20.41 in
T= 168.1 lb
Explanation:
From diagram
Total force balance in vertical direction
T= 89.1 + 79 lb
T= 168.1 lb
Now taking moment about point m
Mm= 0 Because system is in equilibrium position
79 x 43.45 = T x y
Now by putting the value of T
79 x 43.45 = 168.1 x y
y = 20.41 in
So the cable attached at distance of 20.41 in from the mid point of bar.
Tension in the cable = 168.1 lb
A particle is moving along a straight line and has an
acceleration of kV^2 m/s^2, where V is the velocity of the
particle. At time t=0, the velocity of the particle = 4m/s, and the
time t=30s the velocity = 26m/s and displacement at time t = Dt
metres. Derive expressions for both velocity and displacement as a
function of time t.
Answer with Explanation:
By definition of acceleration we have
[tex]a=\frac{dv}{dt}[/tex]
Given [tex]a=kv^{2}[/tex], using this value in the above equation we get
[tex]kv^2=\frac{dv}{dt}\\\\\frac{dv}{v^{1}}=kdt[/tex]
Upon integrating on both sides we get
[tex]\int \frac{dv}{v^2}=\int kdt\\\\-\frac{1}{v}=kt+c[/tex]
'c' is the constant of integration whose value can be found out by putting value of 't' = 0 and noting V =4 m/s
Thus [tex]c=-\frac{1}{4}[/tex]
the value of 'k' can be found by using the fact that at t= 30 seconds velocity = 26 m/s
[tex]\frac{-1}{26}=k\times 30-\frac{1}{4}\\\\\therefore k=\frac{\frac{-1}{26}+\frac{1}{4}}{30}\\\\k=7.05\times 10^{-3}[/tex]
Hence the velocity as a function of time is given by
[tex]v(t)=\frac{-1}{7.05\times 10^{-3}t-\frac{1}{4}}[/tex]
By definition of velocity we have
[tex]v=\frac{dx}{dt}[/tex]
Making use of the obtained velocity function we get
[tex]\frac{dx}{dt}=\frac{-1}{kt-\frac{1}{4}}\\\\\int dx=\int \frac{-dt}{kt-\frac{1}{4}}\\\\x(t)=\frac{-1}{7.05\times 10^{-3}}\cdot ln(7.05\times 10^{-3}t-\frac{1}{4})+x_o[/tex]
here [tex]x_o[/tex] is the constant of integration
A battery is an electromechanical device. a)- True b)- False
Answer:
b)False
Explanation:
A battery is a device which store the energy in the form of chemical energy.And this stored energy is used according to the requirement.So battery is not a electromechanical device.Because it does have any mechanical component like gear ,shaft flywheel etc.
A flywheel is known as mechanical battery because it stored mechanical energy and supply that energy when more energy is required.Generally fly wheel is used during punching operation.
Thermosets burn upon heating. a)-True b)- false?
Answer:
true
Explanation:
True, there are several types of polymers, thermoplastics, thermosets and elastomers.
Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.
Because of the above, thermosetting polymers burn when heated.
A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 constant. The process begins with p1 5 15 lbf/in.2, 1 5 1.25 ft3/lb and ends with p2 5 53 lbf/in.2, 2 5 0.5 ft3/lb. Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value of n. (c) Sketch Process 1–2 on pressure–volume coordinates.
Answer:
V1=5ft3
V2=2ft3
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5ft3
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= 2ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Which is more detrimental to the tool life, a higher depth of cut or a higher cutting velocity? Why?
Answer:
Cutting velocity.
Explanation:
Velocity of cutting affects more as compare to the depth of cut to life of tool.
As we know that tool life equation
[tex]VT^n=C[/tex]
Where T is the tool life and V is the tool cutting speed and C is the constant.
So from we can say that when cutting velocity increase then tool life will decrease and vice versa.
The life of tool is more important during cutting action .The life of tool is less and then will reduce the production and leads to face difficulty .
Describe a physical meaning of velocity and acceleration.
Answer and Explanation:
Velocity : Velocity gives us information about how much we can travel in a particular interval of time.
Velocity is a rate of change of distance if we have information about velocity and time we can easily calculate how much distance we travel in that particular time.
And if we have information about distance and time then we can find how much velocity we should have to travel that distance in given time
Acceleration : Acceleration gives us information about how the velocity is changes with respect to time
If the velocity is increases with time then there is positive acceleration and if velocity is decreases with time then it is negative acceleration.
The time factor for a doubly drained clay layer
undergoingconsolidation is 0.2
a. What is the degree of consolidation (Uz) at z/H=0.25,
0.5,and 0.75
b. If the final consolidation settlement is expected to be
1.0m, how much settlement has occurred when the time factor is 0.2
andwhen it is 0.7?
Answer with Explanation:
Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is
[tex]T_v=\frac{\pi }{4}(\frac{U}{100})^2[/tex]
Solving for 'U' we get
[tex]\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%[/tex]
Since our assumption is correct thus we conclude that degree of consolidation is 50.46%
The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2
i)[tex]\frac{z}{H}=0.25=U=0.71[/tex] = 71% consolidation
ii)[tex]\frac{z}{H}=0.5=U=0.45[/tex] = 45% consolidation
iii)[tex]\frac{z}{H}=0.75U=0.3[/tex] = 30% consolidation
Part b)
The degree of consolidation is given by
[tex]\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm[/tex]
Thus a settlement of 50.46 centimeters has occurred
For time factor 0.7, U is given by
[tex]T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59[/tex]
thus consolidation of 85.59 % has occured if time factor is 0.7
The degree of consolidation is given by
[tex]\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm[/tex]
What is the significance of Saint Venant's principle?
Answer:
While calculating the stresses in a body since we we assume a constant distribution of stress across a cross section if the body is loaded along the centroid of the cross section , this assumption of uniformity is assumed only on the basis of Saint Venant's Principle.
Saint venant principle states that the non uniformity in the stress at the point of application of load is only significant at small distances below the load and depths greater than the width of the loaded material this non uniformity is negligible and hence a uniform stress distribution is a reasonable and correct assumption while solving the body for stresses thus greatly simplifying the analysis.
A rotating cup viscometer has an inner cylinder diameter of 2.00 in., and the gap between cups is 0.2 in. The inner cylinder length is 2.50 in. The viscometer is used to obtain viscosity data on a Newtonian liquid. When the inner cylinder rotates at 10 rev/min, the torque on the inner cylinder is measured to be 0.00011 in-lbf. Calculate the viscosity of the fluid. If the fluid density is 850 kg/m^3, calculate the kinematic viscosity
Answer:
The dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.
Explanation:
Step1
Given:
Inner diameter is 2.00 in.
Gap between cups is 0.2 in.
Length of the cylinder is 2.5 in.
Rotation of cylinder is 10 rev/min.
Torque is 0.00011 in-lbf.
Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.
Step2
Calculation:
Tangential force is calculated as follows:
T= Fr
[tex]0.00011 = F\times(\frac{2}{2})[/tex]
F = 0.00011 lb.
Step3
Tangential velocity is calculated as follows:
[tex]V=\omega r[/tex]
[tex]V=(\frac{2\pi N}{60})r[/tex]
[tex]V=(\frac{2\pi \times10}{60})\times1[/tex]
V=1.0472 in/s.
Step4
Apply Newton’s law of viscosity for dynamic viscosity as follows:
[tex]F=\mu A\frac{V}{y}[/tex]
[tex]F=\mu (\pi dl)\frac{V}{y}[/tex]
[tex]0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}[/tex]
[tex]\mu =1.3374\times 10^{-6}[/tex]lb-s/in².
Step5
Kinematic viscosity is calculated as follows:
[tex]\upsilon=\frac{\mu}{\rho}[/tex]
[tex]\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}[/tex]
[tex]\upsilon=1.4012\times 10^{-3}[/tex] in2/s.
Thus, the dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.
Ductility (increases/decreases/does not change) with temperature.
Answer:
Increases
Explanation:
Ductility:
Ductility is the property of material to go permanent deformation due to tensile load.In other words the ability of material to deform in wire by the help of tensile load.
When temperature is increase then ductility will also increases.And when temperature decreases then the ductility will also decreases.As we know that at very low temperature material become brittle and this is know as ductile brittle transition.
The torque required to drive a 40 mm diameter power screw having double square threads with a pitch of 6 mm is 1.0 kN-m. The friction for the thread is 0.10. Is the condition self-locking? Show your calculation.
Answer:
This is self locking.
Explanation:
Given that
Diameter d= 40 mm
Pitch (P)= 6 mm
Torque T= 1 KN.m
Friction(μ) = 0.1
Thread is double square.
Condition for self locking
[tex]\mu >tan\theta[/tex]
[tex]tan\theta=\dfrac{L} {\pi d}[/tex]
L= n x P
N= number of start
Here given that double start so n=2
L=2 x 6 = 12 mm
[tex]tan\theta=\dfrac{L} {\pi d}[/tex]
[tex]tan\theta=\dfrac{12} {\pi \times 40}[/tex]
[tex]tan\theta=0.095[/tex] (μ = 0.1)
[tex]\mu >tan\theta[/tex]
So from we can that this is self locking.
For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 316 mm2 without plastic deformation? (b) If the original specimen length is 128 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Answer:
a) 93.852 kN
b) 128.043 mm
Explanation:
Stress is load over section:
σ = P / A
If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:
P = σ * A
316 mm^2 = 3.16*10^-4
P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN
The stiffness of the specimen is:
k = E * A / l
k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m
Hooke's law:
x' = x0 * (1 + P/k)
x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm
One horsepower is the ability to lift 550 pounds a height of one foot in one second or one horsepower is equal to 550ft-lb/s. How many ft-lb/s are there in 32 horsepower?
Answer:
32 horsepower will be equal to 17600 ft-lb/sec
Explanation:
We have given 1 horsepower = 550 ft-lb/sec
We have to convert 32 horsepower into ft-lb/sec
As we know that 1 horsepower is equal to 550 ft-lb/sec
So for converting horsepower to ft-lb/sec we have multiply with 550
We have given 32 horse power
So [tex]32\ horsepower =32\times 550=17600ft-lb/sec[/tex]
So 32 horsepower will be equal to 17600 ft-lb/sec
Water enters a tank from two pipes, one with a flow rate of 0.3 kg/s and the other with a flow rate of 0.1 kg / s. The tank has a small hole through which water leaks out at a rate of 0.03 kg / s. If the tank initially contained 40 kg of water, how much will it have after 2 minutes?
Answer:
total amount of water after 2 min will be 84.4 kg/s
Explanation:
Given data:
one tank inflow = 0.1 kg/s
2nd tank inflow = 0.3 kg/s
3rd tank outflow = 0.03 kg/s
Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s
From third point, outflow is 0.03 kg/s
Therefore, resultant in- flow = 0.4 - 0.03
Resultant inflow is = 0.37 kg/s
Tank has initially 40 kg water
In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg
So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws 12.5 hp for 16 h/day, five days per week. Compute the energy used by the mo- tor for one year. Express the result in ft.lb and W.h
The energy used by the motor for one year is 102,313,596,240 ft.lb/year, expressed in both ft.lb and W.h.
1 horsepower (hp) is approximately equal to 745.7 watts (W).
So,[tex]12.5 hp * 745.7 W/hp[/tex]
= 9312.5 W.
The motor draws 9312.5 watts for 16 hours per day.
So, [tex]9312.5 W * 16 hours/day[/tex]
= 148,200 W.h/day.
Multiply the daily energy by the number of days per year:
The motor operates for 5 days per week, so in one year (considering 52 weeks), it operates for 5 days/week × 52 weeks/year = 260 days/year.
[tex]148,200 W.h/day *260 days/year[/tex]
= 38,532,000 W.h/year.
To express the result in ft.lb, we need to convert the energy from watt-hours to foot-pounds:
1 watt-hour (Wh) is approximately equal to 2655.22 foot-pounds (ft.lb).
So, [tex]38,532,000 W.h/year * 2655.22 ft.lb/W.h[/tex]
≈ 102,313,596,240 ft.lb/year.
Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Consider a drag of 12 kN on the shuttle. Calculate the work done by the shuttle engine.
Answer:
work done = 48.88 × [tex]10^{9}[/tex] J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × [tex]10^{9}[/tex] J
A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3m3. A constant pressure process gives 18 kJ of work out. What is the final temperature? You may assume ideal gas.
Answer:
1160 K.
Explanation:
Given that
Initial
Pressure P =600 KPa
Temperature T =290 K
Volume V =0.01 [tex]m^3[/tex]
If we assume that air is s ideal gas the
P V = mRT
R=0.287 KJ/kg.k
now by putting the values in above equation
600 x 0.01 = m x 0.287 x 290
m=0.07 kg
The work out at constant pressure given as
[tex]w=P(V_2-V_1)[/tex]
[tex]18=600(V_2-0.01)[/tex]
[tex]V_2=0.04\ m^3[/tex]
At constant pressure
[tex]\dfrac{T_2}{T_1}=\dfrac{V_2}{V_1}[/tex]
[tex]\dfrac{T_2}{290}=\dfrac{0.04}{0.01}[/tex]
[tex]T_2=1160\ K[/tex]
So the final temperature is 1160 K.
If the original length of a specimen is L0 = 10"" and new length of the specimen after applied load is L = 12.5"". The value of true strain is: a) 0.5 b) 0.25 c) 0.223 d) 0.4
Answer:
The correct answer is option 'b':0.25
Explanation:
By definition of strain we have
[tex]\epsilon =\frac{L_f-L_o}{L_o}[/tex]
where
[tex]\epsilon [/tex] is the strain
[tex]L_o[/tex] is the original length of specimen
[tex]L_f[/tex] is the elongated length of specimen
Applying the given values we get
[tex]\epsilon =\frac{12.5''-10''}{10''}=0.25[/tex]
Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158 mm, what are the temperature and velocity at the exit?
Answer:
[tex]v_2 = 160.23 m/s[/tex]
[tex]T_2 = 475.797 k[/tex]
Explanation:
given data:
Diameter =[tex] d_1 = 200mm[/tex]
[tex]t_1 =195 degree[/tex]
[tex]p_1 =500 kPa[/tex]
[tex]v_1 = 100m/s[/tex]
[tex]p_2 = 85kPa[/tex]
[tex]d_2 = 158mm[/tex]
from continuity equation
[tex]A_1v_1 = A_2v_2[/tex]
[tex]v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}[/tex]
[tex]v_2 = \frac{d_2v_1}{d_2^2}[/tex]
[tex]v_2 = [\frac{d_1}{d_2}]^2 v_1[/tex]
[tex]= [\frac{0.200}{0.158}]^2 \times 100[/tex]
[tex]v_2 = 160.23 m/s[/tex]
by energy flow equation
[tex]h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w[/tex]
[tex]z_1 =z_2[/tex] and q =0, w =0 for nozzle
therefore we have
[tex]h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2} [/tex]
[tex]dh = \frac{1}{2} (v_1^2 -v_2^2)[/tex]
but we know dh = Cp dt
hence our equation become
[tex]Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)[/tex]
[tex]Cp (T_2 -T_1) = 7836.94[/tex]
[tex](T_2 -T_1) = \frac{7836.94}{1.005*10^3}[/tex]
[tex](T_2 -T_1) = 7.797 [/tex]
[tex]T_2 = 7.797 +468 = 475.797 k[/tex]
At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates such that for each 5.7 lbf applied, the spring elongates one inch. If local acceleration of gravity is 32.2 ft/s2, what distance, in inches, did the spring elongate?
Answer:
x=2.19in
Explanation:
This is the equation that relates the force and displacement of a spring
F=Kx
m=mass=12.5lbx1slug/32.14lb=0.39slug
F=mg=0.39*32.2=12.52Lbf
then we calculate the spring count in lbf / ft
K=F/x
K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft
Finally we calculate the displacement with the initial equation
X=F/k
x=12.52/68.4=0.18ft=2.19in
In this exercise we want to calculate how much the spring was elogate, for this we have to be:
the spring stayed x=2.19in elogante
organizing the information given in the statement we have that:
mass of 12.5 lb force each 5.7 lbf appliedacceleration of gravity is 32.2 ft/sRecalling the basic equation of the spring we find that:
[tex]F=Kx[/tex]
Where:
F is the applied forceK is the spring constantX is how much the spring has been elongatedSo calculating the force we have:
[tex]F=mg\\=0.39*32.2\\=12.52[/tex]
Putting the value of the force in the given formula:
[tex]K=F/x\\K=5.7/1\\=5.7\\X=F/k=12.52/68.4\\=0.18ft\\=2.19in[/tex]
See more about spring at brainly.com/question/4433395
Derive the following conversion factors: (a) Convert a pressure of 1 psi to kPa (b) Convert a vol ume of 1 liter to gallons (c) Convert a viscosity of 1 lbf .s/ft^2 to N s/m^2
Answer:
a)6.8 KPa
b)0.264 gallon
c)47.84 Pa.s
Explanation:
We know that
1 lbf= 4.48 N
1 ft =0.30 m
a)
Given that
P= 1 psi
psi is called pound force per square inch.
We know that 1 psi = 6.8 KPa.
b)
Given that
Volume = 1 liter
We know that 1000 liter = 1 cubic meter.
1 liter =0.264 gallon.
c)
[tex]1\ \frac{lb.s}{ft^2}=47.84\ \frac{Pa.s}{ft^2}[/tex]
A mass of 2 kg is hanging to a spring of 200 n/m spring constant vertically. Calculate the period of the period if the spring is set in simple harmonic motion.
Answer:
The period is 0.628s
Explanation:
As the system mass-spring is in simple harmonic motion, the period is given by the equation:
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where m is the mass of 2kg and k is the spring constant [tex]200\frac{N}{m}[/tex]
Replacing the values in the equation, we have:
[tex]T=2\pi \sqrt{\frac{2Kg}{200\frac{N}{m}}}[/tex]
And finally we find the value of the period, that is:
[tex]T=0.628s[/tex]
What is the difference between pound-mass and pound-force?
Answer:
Pound mass is the unit of mass and which is used in United states customary to describe the mass Slug is also a unit of mass .Pound mass is lbm. But on the other hand pound force is the unit of force.Pound force is lbf.
In SI unit ,the unit of mass is kilogram(kg) and the unit of force is Newton(N)
In CGS system the unit of mass is gram and unit of force is dyne.
Calculate the efficiency of a Carnot Engine working between temperature of 1200°C and 200°C.
Answer:
efficiency = 0.678
Explanation:
First we have to change temperatures from °C to K (always in thermodynamics absolute temperature is used).
[tex]\text{hot body temperature,}T_H = 1200 \circC + 273.15 = 1473.15 K[/tex]
[tex] \text{cold body temperature,} T_C = 200 \circC + 273.15 = 473.15 K[/tex]
Efficiency [tex] \eta [/tex] of a carnot engine can be calculated only with hot body and cold body temperature by
[tex] \eta = 1 - \frac{T_C}{T_H}[/tex]
[tex] \eta = 1 - \frac{473.15 K}{1473.15 K}[/tex]
[tex] \eta = 0.678[/tex]
The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer height of 50 ft and an outer diameter of 4.6 ft. The tanks are made of 0.1 ft thick steel (steel = 499 lbm/ft?) and store air at a maximum pressure of 2500 psi at -10 °F. How much load must the support structure at the base of the tanks carry?
Answer:
179000 lb
Explanation:
The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.
The specific weight of water is ρw = 62.4 lbf/ft^3
These tanks are thin walled because
D / t = 4.6 / 0.1 = 46 > 10
To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:
A = 2 * π/4 * D^2 + π * D * h
The steel volume is:
V = A * t
The specific weight is
ρ = δ * g
ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3
The weight of the steel tank is:
Ws = ρs * V
Ws = ρs * A * t
Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t
Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb
The weight of water can be approximated with the volume of the tank:
Vw = π/4 * D^2 * h
Ww = ρw * π/4 * D^2 * h
Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb
Wt = Ws + Ww = 37700 + 51800 = 89500 lb
Assuming the support holds both tanks
2 * 89500 = 179000 lb
The support must be able to carry 179000 lb
A steel band blade, that was originally straight, passes
over8-in.-diameter pulleys when mounted on a band saw. Determine
themaximum stress in the blade, knowing that it is 0.018 in. thick
and0.625 in. wide. Use E = 29 x 106 psi.
Answer:
[tex]\sigma = 65.25\ ksi[/tex]
Explanation:
given data:
D = 8 inch
R =4 inch
thickness = 0.018 inch
width = 0.625 inch
[tex]E =29*10^6 psi[/tex]
from bending equation we know that
[tex]\frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R}[/tex]
[tex]\sigma = \frac{Ey}{R}[/tex]
Where y represent distance from neutral axis
[tex]y = \frac{t}[2}[/tex]
[tex]y = \frac{0.018}{2}[/tex]
y = 0.009inch
[tex]\sigma = \frac{29*10^6*0.009}{4}[/tex]
[tex]\sigma = 65250\ psi[/tex]
[tex]\sigma = 65.25\ ksi[/tex]
If a pendulum is 10m long, (a) what is the natural frequency and the period of vibration on the earth, where the free-fall acceleration is 9.81 m/s^2 and (b) what is the natural frequency and the period of vibration on the moon, where the free-fall acceleration is 1.67 m/s^2?
Answer:
(a) Natural frequency = 0.99 rad/sec (b) 0.4086 rad/sec
Explanation:
We have given length of pendulum = 10 m
(a) Acceleration due to gravity [tex]=9.81m/sec^2[/tex]
Time period of pendulum is given by [tex]T=2\pi\sqrt{\frac{L}{g}}[/tex], L is length of pendulum and g is acceleration due to gravity
So [tex]T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{9.81}}=6.34sec[/tex]
Natural frequency is given by [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6.34}=0.99rad/sec[/tex]
(b) In this case acceleration due to gravity [tex]g=1.67m/sec^2[/tex]
So time period [tex]T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{1.67}}=15.3674sec\[/tex]
Natural frequency [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{15.36}=0.4086rad/sec[/tex]