Determine the weight of a 2,838 kg car in lbs. Round to the nearest 1lb.

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.

Answer:

250 kpsi

Explanation:

Given:

Width of the specimen, w = 0.45 in

Thickness of the specimen, t = 0.20 in

length of the specimen between supports, L = 2.5 in

Failure load, F = 1200 lb

Now,

The transverse rupture strength [tex]\sigma_t=\frac{1.5FL}{wt^2}[/tex]

on substituting the respective values, we get

[tex]\sigma_t=\frac{1.5\times1200\times2.5}{0.45\times0.2^2}[/tex]

or

[tex]\sigma_t=250,000\ psi\ =\ 250 kpsi[/tex]

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]

Answer:

1) The initial velocity must be 22 feet/sec.

2) The time required to change the velocity is 8 seconds.

Explanation:

This problem can be solved using third equation of kinematics

According to the third equation of kinematics we have

[tex]v^{2}=u^{2}+2as[/tex]

where

'v' is the final velocity of object

'u' is the initial velocity of object

'a' is acceleration of the object

's' is the distance in which this velocity change is brought

Since the final velocity is 45 mph converting this to feet per seconds we get

45 mph =66 feet/sec

Applying Values in the above equation and solving for 'u' we get

[tex]66^{2}=u^{2}+2\times 5.5\times 352\\\\u^{2}=484\\\\\therefore u=22feet/sec[/tex]

2)

The time required to attain this velocity can be found using first equation of kinematics as

[tex]v=u+at[/tex]

where

't' is the time in which the velocity change occurs

'v', 'u', 'a' have the usual meaning

Thus applying given values we get

[tex]66=22+5.5\times t\\\\\therefore t=\frac{66-22}{5.5}=8seconds[/tex]

Applying the given values we get

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]

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.

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

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.

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°".

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

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = [tex]\frac{\pi }{4} d^2[/tex]    ..................1

here d is diameter

put the value in equation 1

area = [tex]\frac{\pi }{4} 0.15^2[/tex]

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = [tex]\frac{Fnet}{g}[/tex]

mass  = [tex]\frac{350}{9.8}[/tex]

mass = 35.7 kg

so 35.7 kg lid we put

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]

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.

Answer:

Final velocity will be 36.11 m/sec

Time required by ball to heat the ground = 1.297 sec

Explanation:

We have given height = 50 feet

Initial velocity u = 18 ft/sec

Acceleration due to gravity [tex]g=32.8ft/sec^2[/tex]

We have to find final velocity v

From third equation of motion

[tex]v^2=u^2+2gh[/tex]

So [tex]v^2=18^2+2\times 32.8\times 50=3543.82[/tex]

v = 59.53m/sec

From first equation of motion we know that

v= u+gt

So [tex]59.53=18+32.8\times t[/tex]

t = 1.297 sec

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

Answer:

Mass flow in kg /sec will be 1.511 kg/sec

Explanation:

We have given mass flow = 200 lbm/min

We have to convert lbm/min into kg /sec

We know that 1 lbm = 0.4535 kg

So for converting lbm to kg we have multiply with 0.4535

So 200 lbm = 200×0.4535 =90.7 kg

Now we know that 1 minute = 60 sec

So 200 lbm/min [tex]=\frac{200\times 0.4535kg}{60sec}=1.511kg/sec[/tex]

So mass flow in kg /sec will be 1.511 kg/sec

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

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]

Answer:

The answer is "b" 3.5W and 37.5w.

Explanation:

Domestic use normally refers to refrigeration equipment that is not subject to high thermal loads, for example, refrigerators should only extract heat from food, and domestic air conditioners should only extract heat from small spaces and a number reduced people.

Answer:

Star

Explanation:

When all hosts are connected to a central hub it is known as a star topology.

The advantages of the star network is that it is very easy to add new devices, a failure in a host will not cause problems on the rest of the network, it is appropriate for large networks and works well under heavy loads,

The disadvantages are that a failure of the central hub brings the whole networks down and it is expensive due to the amount of cables needed.

Answer:

power = 20943.95 watts

power = 28.086 horsepower

Explanation:

given data

torque = 200 N

speed = 1000 rpm

to find out

What is the mechanical power in watts and horse power

solution

we know that mechanical power formula that is

power = torque × speed   ...................1

here we have given both torque and speed

we know speed = 1000 rpm = [tex]\frac{2* \pi *1000}{60}[/tex] = 104.66 rad/s

so put here value in equation 1

power = 200 × 104.719

power = 20943.95 watts

and

power = [tex]\frac{20943.95}{745.7}[/tex]

power = 28.086 horsepower

Answer:

Part 1) Power required for motor = 20944 watts.

Part 2) Power required for motor in Horsepower equals = 28.075H.P

Explanation:

Power is defined as the rate of consumption of energy. For rotational motion power is calculated as

[tex]Power=Torque\times \omega[/tex]

where,

[tex]\omega [/tex] is the angular speed of the motor.

Since the rotational speed of the motor is given as 1000 rpm, the angular speed is calculated as

[tex]\omega =\frac{N}{60}\times 2\pi[/tex]

where,

'N' is the speed in rpm

Applying the given values we get

[tex]\omega =\frac{1000}{60}\times 2\pi=104.72rad/sec[/tex]

hence the power equals

[tex]Power=200\times 104.72=20944Watts[/tex]

Now since we know that 1 Horse power equals 746 Watts hence 20944 Watts equals

[tex]Power_{H.P}=\frac{20944}{746}=28.075H.P[/tex]

Answer:

18.7 mm

Explanation:

Fourier's law for heat conduction on a plate is:

q = -k/t * ΔT

Where

q: heat conducted per unit of time and surface

k: thermal conductivity

t: thickness of the plate

ΔT: temperature difference

Rearranging:

t = -k/q * ΔT

t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (q is negative because it is heat leaving the plate)

The maximum thickness of the concrete slab is 18.7 mm.

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.

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.

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

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

Answer:

r=0.228m

Explanation:

The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases

PV=nRT

where

P=pressure =5bar=500.000Pa

V=volume

n=moles=10

R = universal constant for ideal gases = 8.31J / (K.mol)

T=temperature=80F=299.8K

solvig For V

V=(nRT)/P

[tex]V=(\frac{(10)(8.31)(299.8)}{500000} )\\V=0.0498m^3[/tex]

we know that the volume of a sphere is

[tex]V=\frac{4\pi r^3}{3} \\[/tex]

solving for r

[tex]r=\sqrt[3]{ \frac{3 V}{4\pi } }[/tex]

solving

[tex]r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m[/tex]

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

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

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.  

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.