the force acting on a particle of mass m is given by
F=kvx
in which k is a positive constant. The particle passes
throughthe origin with speed Vo at time t=0. Find x as a function
oft.

Answers

Answer 1

Answer with Explanation:

From newton's second law the acceleration produced by a force on a mass 'm' is given by

[tex]Acceleration=\frac{Force}{Mass}[/tex]

Applying the given values in the above equation we get

[tex]Acceleration=\frac{kvx}{m}[/tex]

Also we know that acceelration of a particle can ve mathem,atically written as

[tex]a=\frac{v\cdot dv}{dx}[/tex]

Applying the given values in the above equation we get

[tex]\frac{kvx}{m}=\frac{v\cdot dv}{dx}\\\\\Rightarrow {kx}\cdot dx=m\cdot dv\\\\\int kxdx=\int mdv\\\\\frac{kx^2}{2}=mv-c[/tex]

'c' is the constant of integration

whose value is found that at x =0 v= [tex]v_o[/tex]

Thus

[tex]c={mv_o}[/tex]

Thus the velocity as a function of position is

[tex]v=\frac{1}{m}(\frac{kx^2}{2}+c)[/tex]

Now  by definition of velocity we have

[tex]v=\frac{dx}{dt}[/tex]

Using the function of velocity in the above relation we get

[tex]\frac{dx}{kx^{2}+\sqrt{2c}}=\frac{dt}{2m}\\\\\int \frac{dx}{(\sqrt{k})^2x^{2}+(\sqrt{2c})^2}=\int \frac{dt}{2m}\\\\\frac{1}{\sqrt{2kc}}\cdot tan^{-1}(\frac{(\sqrt{k})x}{\sqrt{2c}})=\frac{t}{2m}+\phi \\\\[/tex]

where

[tex]\phi [/tex] is constant of integration

Now it is given that at t = 0 ,x = 0

thus from the above equation of position and time we get [tex]\phi =0[/tex]

Thus the position as a function of time is

[tex]x(t)=\sqrt{\frac{2c}{k}}\cdot tan(\frac{kct}{\sqrt{2}m})[/tex]

where c=[tex]mv_o[/tex]


Related Questions

What is mass flow of 200 lbm/min in kg/sec?

Answers

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

Consider two closed systems A and B. System A contains 300 kJ of thermal energy at 20C, whereas system B contains 200kJ of thermal energy at 50oC. Now the systems are brought into contact with each other. Determine the direction of any heat transfer between the two systems and explain your answer.

Answers

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.

An electric motor supplies 200 N·m of torque to a load. What is the mechanical power supplied to the load if the shaft speed is 1000 rpm? Express the result in watts and horsepower.

Answers

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]

Higher molecular weight results in worst mechanical properties. a)-True b)- false?

Answers

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.

If a plus sight of 12.03 ft is taken on BM A, elevation 312.547 ft, and a minus sight of 5.43 ft is read on point X, calculate the HI and the elevation of point X.

Answers

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

You want a potof water to boil at 105 celcius. How heavy a
lid should youput on the 15 cm diameter pot when Patm =
101kPa?

Answers

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

a piston moves a 25kg hammerhead vertically down 1m from rest to a
velocity of 50m/s in a stamping machine.

what is the total change in energy of the hammerhead?

Answers

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

A closed, rigid tank contains 2 kg of water initially at 80 degree C and a quality of 0.6. Heat transfer occurs until the tank contains only saturated vapor at a higher pressure. Kinetic and potential energy effects are negligible. For the water as the system, determine the amount of energy transfer by heat, in kJ.

Answers

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.

A heating cable is embedded in a concrete slab for snow melting. The heating cable is heated electrically with joule heating to provide the concrete slab with a uniform heat of 1200 W/ m2. The concrete has a thermal conductivity of 1.4 W/m⋅K. To minimize thermal stress in the concrete, the temperature difference between the heater surface (T1) and the slab surface (T2) should not exceed 16°C (2015 ASHRAE Handbook—HVAC Applications, Chapter. 51). Formulate the temperature profile in the concrete slab, and determine the thickness of the concrete slab (L) so that T1 − T2 ≤ 16°C.

Answers

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.

What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?

Answers

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.  

A tow truck is using a cable to pull a car up a 15 degree hill. If the car weighs 4000 lbs and the cable has a diameter of .75 inches, find the stress in the cable when the truck comes to a stop while on the hill. Ignore friction between the car and the pavement.

Answers

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]

The charpy test determines?

Answers

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.

A certain working substance receives 100 Btu reversibly as heat at a temperature of 1000℉ from an energy source at 3600°R. Referred to as receiver temperature of 80℉, calculate: A. the available energy of the working substance ( 63 Btu ) B. the available portion of the 100 Btu added at the source temperature ( 85 Btu ) C. the reduction in available energy between the source temperature and the 1000℉ temperature ( 22 Btu )

Answers

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

The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.

Answers

Answer:

Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]

Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]

[tex]g_{s} = 8.79076[/tex]

[tex]w_{w} = 9782.36 N/m^{3}[/tex]

Given:

Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]

Mass of a can of pop , m = 0.369 kg

Weight of an empty can, W = 0.153 N

Solution:

Now, weight of a full can pop, W

W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]

Now weight of the pop in can is given by:

w = W' - W = 3.616 - 0.513 = 3.103 N

Now,

The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]

[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]

Now, density of the pop:

[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]

[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]

Now,

Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]

where

[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]

Now, for water at [tex]20^{\circ}c[/tex]:

Specific density of water = [tex]998.2 kg/m^{3}[/tex]

Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]

Specific weight of water at [tex]20^{\circ}c[/tex]:

[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]

A car is accelerated 5.5 ft/s^2. Calculate the initial velocity v, the car must have if it is to attain a final velocity v of 45 mph in a distance of 352 ft. Compute also the time t required to attain that final velocity.

Answers

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

If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew point? What is the humidity ratio? What is the enthalpy?

Answers

Answer:

Relative humidity 48%.

Dew point 74°F

humidity ratio 118 g of moisture/pound of dry air

enthalpy 41,8 BTU per pound of dry air

Explanation:

You can get this information from a Psychrometric chart for water, like the one attached.

You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.

To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F

for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air

For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air

When all network cables connect to one central point the network topology is typically referred to as a(n)_______

Answers

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.

A carpenter uses a hammer to strike a nail. Approximate the hammer's weight of 1.8lbs, as being concentrated at the head, and assume that at impact the head is traveling in the -j direction. If the hammer contacts the nail at 50mph and the impact occurs over 0.023 seconds, what is the magnitude of the average force exerted by the nail of the hammer?

Answers

Answer:

The average force F exherted by the nail over the hammer is 178.4 lbf.

Explanation:

The force F exherted by the nail over the hammer is defined as:

F = |I|/Δt

Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:

I = ΔP = Pf - Pi

Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:

P = m.v  

Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:

ΔP =  - Pi = -m.vi

The initial velocity is given as 50 mph and we will expressed in ft/s:

vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s

By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:

|I|= |ΔP|= |-m.vi| = 1.8 lb  * 73.3 ft/s = 132 lb.ft/s

Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:

F =  132 lb.ft/s / 0.023 s = 5740 lb.ft/s^2  

Expressing in lbf:

F = 5740 lb.ft/s^2 * 0.031 lbf/lb.ft/s^2  =  178.4 lbf

A large water jet with a discharge of 2m^3 /s rises 90m above the ground. The exit nozzle diameter to achieve this must be. (a) 0.246m (b) 0.318m (c) 1.3m (d) 0.052m (e) None of the above

Answers

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

[tex]v=\sqrt{2gh}[/tex]

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

[tex]v=\sqrt{2\times 9.81\times 90}=42.02m/s[/tex]

now we know the relation between discharge and velocity as dictated by contuinity equation is

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

Applying values in the above equation and solving for area we get

[tex]Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}[/tex]

The circular area is related to diameter as

[tex]Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m[/tex]

Thus the diameter of the nozzle is 0.246 meters

A brittle intermetallics specimen is tested with a bending test. The specimen's width 0.45 in and thickness 0.20 in. The length of the specimen between supports 2.5 in. Determine the transverse rupture strength if failure occurs at a load 1200 lb.

Answers

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]

A spherical tank is being designed to hold 10 moles of carbon dioxide gas at an absolute pressure of 5 bar and a temperature of 80°F. What diameter spherical tank should be used? The molecular weight of methane is 44 g/mole.

Answers

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]

Various factors to be considered in deciding the factor of safety?

Answers

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.

a) What is the Damkohler number? b) What is the significance of a system with a low Damkohler number?

Answers

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.

A Carnot heat engine operates between 1000 deg F and 50 deg F, producing 120 BTU of work. What is the heat input to the engine?

Answers

Answer:

184.6 BTU

Explanation:

The thermal efficiency for a Carnot cycle follows this equation:

η = 1 - T2/T1

Where

η: thermal efficiency

T1: temperature of the heat source

T2: temperature of the heat sink

These temperatures must be in absolute scale:

1000 F = 1460 R

50 F = 510 R

Then

η = 1 - 510/1460 = 0.65

We also know that for any heat engine:

η = L / Q1

Where

L: useful work

Q1: heat taken from the source

Rearranging:

Q1 = L / η

Q1 = 120 / 0.65 = 184.6 BTU

In this exercise we will deal with the Carnot cycle and calculate how much heat was initially placed, so we have that:

The heat input to the engine was 184.6 BTU

How does the Carnot Cycle work?

The theoretical Carnot cycle is formed by two isothermal transformations and two adiabatic transformations. One of the isothermal transformations is used for the temperature of the hot source, where the expansion process takes place, and the other for the cold source, where the compression process takes place.

The thermal efficiency for a Carnot cycle follows this equation:

[tex]\eta= 1 - T_2/T_1[/tex]

Where:

η: thermal efficiencyT1: temperature of the heat sourceT2: temperature of the heat sink

These temperatures must be in absolute scale:

[tex]\eta = 1 - 510/1460 = 0.65[/tex]

Rearranging:

[tex]Q_1 = L / \eta\\Q_1 = 120 / 0.65 = 184.6 BTU[/tex]

See more about carnot cycle at brainly.com/question/21126569

a baseball is thrown downward from a 50 ft tower with an
initialspeed of 18 ft/s.
what is the speed when the ball hits the ground and the time
oftrave?

Answers

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

The flow curve for a certain metal has parameters: strain-hardening
exponent is 0.22 and strength coefficient is54,000
lb/in2 . Determine:
a) the true stress at a true strain = 0.45
b) the true strain at a true stress = 40,000
lb/in2.

Answers

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]

Define drag and lift forces.

Answers

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.

Domestic units have power input ratings of between (a) 35 W and 375 W (b) 3.5 and 37.5 W (c) 350W and 375 W

Answers

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.

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C, what will be the percentage change in its volume? If the vat has a diameter of 2 m, how much will the water level rise due to this temperature increase?

Answers

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

As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth h1 = 42 mm and specific gravity S = 0.8. The reservoir is connected to a round tube of diameter D2 = 5 mm and oil rises in the tube to height h2. Find h2. Assume the oil in the tube is open to atmosphere and neglect the mass of the piston.

Answers

Final answer:

Use Pascal's law and the hydrostatic pressure formula to solve for the height of the oil column, considering the specific gravity of the oil and the mass applied on the piston.

Explanation:

The question requires the application of principles from fluid mechanics to calculate the height h2 to which an oil rises in a tube, using the specific gravity of the oil and the mass of a load on a piston. Assuming the system is in equilibrium and by using Pascal's law, the pressure applied by the mass onto the piston is transmitted equally throughout the fluid. Therefore, the pressure exerted by the mass on the larger piston is balanced by the hydrostatic pressure of the oil column of height h2 in the smaller tube.

To solve for h2, we can use the formula for hydrostatic pressure P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column. Given the specific gravity S of the oil, we can find its density (ρ) by multiplying S by the density of water (1000 kg/m³). Then, by setting the hydrostatic pressure of the oil column equal to the pressure exerted by the mass and solving for h2, we find the height of the oil in the tube.

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