Yield and tensile strengths and modulus of elasticity with increasing temperature. (increase/decrease/independent)

Answers

Answer 1

Answer:

Decrease

Explanation:

Generally with increasing the temperature the mechanical properties of material decreases.But some materials have exceptions like tempered steel because when temperature increase then young modulus of elasticity of tempered steel increases.

So we can say that when  with increasing temperature the properties of materials Yield and tensile strengths and modulus of elasticity decreases.

Decrease

Answer 2

Answer:

Decreases

Explanation:

Yield and tensile strength and modulus of elasticity decreases with increasing temperature. As the temperature is increased most materials decrease in their elasticity.

The modulus of elasticity is proportional to its tensile strength.


Related Questions

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?

Answers

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

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

Answers

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.

Given 10 parts per billion. What is the concentration
inmg/L?

Answers

Answer:

The concentration in mg/L equals 0.01 mg/L

Explanation:

Since the concentration is given as 10 parts per billion

We know that

1 part per billion equals 0.001 mg/L

Thus 10 parts per billion  concentration equals

[tex]10\times 0.001mg/L\\\\=0.01mg/L[/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?

Answers

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]

What does stall mean?

Answers

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.

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?

Answers

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]

Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 mm, determine the minimum flow rate that the pump must provide.

Answers

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

[tex]Q = A \times V[/tex]

where A is Area

[tex]A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2[/tex]

[tex]Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s[/tex]

minimum flow rate provided by pump is 0.02513 m^3/s

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?

Answers

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

To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01

Answers

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

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?

Answers

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

Describe a physical meaning of velocity and acceleration.

Answers

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.

An ideal Otto Cycle with air as the working fluid has a compression ratio of10. The minimum temperature and pressure in the cycle are 25 degrees celsius and 100 kPa respectively. The highest temperature in the cycle is 1000 degrees celsius. Using the PG model for air with properties: k =1.4, c_p = 1 kJ.kg.K, c_v = 0.717 kJ/kg.K, determine the exhaust temperature (temperature at the end of the expansion stroke) in Kelvins.

Answers

Answer:

[tex]T_4[/tex]= 506.79 K

Explanation:

Given that

Minimum temperature [tex]T_1[/tex]= 25 C

Pressure = 100 KPa

Highest temperature [tex]T_3[/tex]= 1000 C

Compression ratio r= 10

We know that for ideal Otto cycle

[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]

Where  [tex]T_3[/tex] is the exhaust temperature.

Now by putting the values

[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]

[tex]\dfrac{1000+273}{T_4}=10^{1.4-1}[/tex]

[tex]T_4[/tex]= 506.79 K

Which is more detrimental to the tool life, a higher depth of cut or a higher cutting velocity? Why?

Answers

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 .

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

Answers

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]

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.

Answers

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 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.

Answers

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

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?

Answers

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/s

Recalling 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 elongated

So 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

Thermosets burn upon heating. a)-True b)- false?

Answers

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.

An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50 km/h? If the vehicle accelerates to 100 km/h in 15 seconds, what is the change in power required?

Answers

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

[tex]Ek=0.5mv^{2}[/tex]

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

[tex]Ek=0.5(1200)(13.89)^2[/tex]

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

[tex]Ek=0.5(1200)(27.78)^2=462963J\\[/tex]

taking into account all of the above the following equation is inferred

ΔE=[tex]\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15}  =23146.916W=23.15kW[/tex]

What is the difference between pound-mass and pound-force?

Answers

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.

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.

Answers

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

Calculate the efficiency of a Carnot Engine working between temperature of 1200°C and 200°C.

Answers

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]

What is the significance of Saint Venant's principle?

Answers

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 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?

Answers

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

Ductility (increases/decreases/does not change) with temperature.

Answers

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.

Starting from position s = 0 m at time t = 0 s, a particle travels on a straight-line path. The particle's velocity is given by the function v = 5 sin (4s2) m/s, where s is in meters and the argument of the sine function is unitless. Find the particle's acceleration when s = 0.25 m and s = 1.25 m.

Answers

Answer:

acceleration is 0.2181 m/s²

acceleration is 27.05 m/s²

Explanation:

given data

s = 0

time t = 0

velocity V = 5 sin (4s²) m/s

to find out

particle's acceleration

solution

we have given velocity = 5 sin(4s²)

and we know here that velocity = [tex]\frac{ds}{dt}[/tex]

so acceleration will be a =   [tex]\frac{dv}{dt}[/tex]

put here velocity v

acceleration =  [tex]\frac{dv}{dt}[/tex]

acceleration =  [tex]\frac{d(5sin(4s^4))}{dt}[/tex]

acceleration = 5 cos4s² × 8s × [tex]\frac{ds}{dt}[/tex]

acceleration =  5 cos4s² × 8s × 5 sin4s²

acceleration = 200 s ×cos4s²  × sin4s²

put here s = 0.25 and s = 1.25

so

acceleration = 200× 0.25 ×cos(4×0.25²)  × sin(4×0.25²)

acceleration = 0.2181 m/s²

acceleration = 200× 1.25 ×cos(4×1.25²)  × sin(4×1.25²)

acceleration = 27.05 m/s²

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?

Answers

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

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.

Answers

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 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.

Answers

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]

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.

Answers

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.

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