The bars of the truss each have a cross-sectional area of 1.25 in2. If the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude P of the loads that can be applied to the truss.

Answers

Answer 1

Answer:

P=25000lbf

Explanation:

For this problem we will use the equation that relates, the effort, the area and the force for an element under normal stress.

σ=P/A

σ=stress=20kSI=20 000 lbf/in ^2

P=force

A=area

solving for P

P=Aσ

P=(20 000 lbf/in ^2)(1.25in^2)

P=25000lbf

Answer 2

Answer:

The maximum magnitude P of the loads that can be applied to the truss = 25,000 Pounds or 111,205.5 Newtons.

Explanation:

In order to calculate the maximum load P, we will make use of the formula: Maximum average stress (20 ksi) = maximum load P ÷ cross-sectional area (1.25 in²)

Make P (the maximum load) the subject of the formula: P = 20 ksi × 1.25 in².

Before moving further, we have to convert the average normal stress (in ksi) to an appropriate unit: The average normal stress = 20 ksi = 20 kip per square inch (kip/in²)

But 1 kip = 1000 Pounds (i.e., 1000 lb)

Therefore, 20 ksi = 20,000 Pounds/in².

Therefore, P (maximum load) = 20,000 pounds/in² × 1.25 in² = 25,000 Pounds = 111,205.5 Newtons (because 1 Pound = 4.44822 Newtons).


Related Questions

A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates with a maximum amplitude of 10 mm. What is its natural frequency?

Answers

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

[tex]V_{max}=\omega A[/tex]

now by putting the values

[tex]V_{max}=\omega A[/tex]

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

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.

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.

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.

Derive the dimensions of specific heat that is defined as the amount of heat required to elevate the temperature of an object of mass 1 kg by 1 degree Celcius.

Answers

Answer:

Dimension of specific heat will be [tex]=L^2T^{-2}\Theta ^{-1}[/tex]

Explanation:

We know that heat [tex]Q=mc\Delta T[/tex], Q is heat generated, m is mass, c is specific heat and [tex]\Delta T[/tex] is temperature difference

From formula we can write [tex]c=\frac{Q}{m\times \Delta T}[/tex]

Now unit of Q is joule or N-m

Newton can be written as [tex]kgm/sec^2[/tex]

So unit of Q is [tex]kgm^2/sec^2[/tex]

For dimension we use M for kg, L for meter(m) ,T for sec and [tex]\Theta[/tex] for temperature

So dimension of Q is [tex]ML^2T^{-2}[/tex]

So dimension of specific heat will be [tex]\frac{ML^2T^{-2}}{M\Theta }=L^2T^{-2}\Theta ^{-1}[/tex]

Give examples of engineering structures which can be modelled as thin walled cylinders.

Answers

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.

The higher the degree of polymerization, the longer the chains will be, which results in a lower chain molecular weight. a)-True b)- false?

Answers

Answer:

b)False

Explanation:

When one or more than one monomer  join to other monomer then they form a chain and this joining of monomers is called degree of polymerization .

Degree of polymerization :

[tex]Degree\ of\ polymerization=\dfrac{Mass\ of\ polymer}{Mass\ of\ monomer}[/tex]

So from above we can say that when chain will become longer then the weight of polymer will increase.

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

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 4-kg-plastic tank that has a volume of 0.2 m^3 is filled with liquid water. Assuming the density of water is 1000 kg/m^3, determine the weight the combined system.

Answers

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]

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.

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

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

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) 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 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

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]

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.

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.

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]

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.

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 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]

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

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.

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]

A force is specified by the vector F= 160i + 80j + 120k N. Calculate the angles made by F with the positive x-, y-, and z-axis.

Answers

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°".

Force and Vector:

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

A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3000 kPa and 25 kPa. The temperature of the steam at the turbine inlet is 400 oC, and the mass flow rate of steam through the cycle is 37 kg/s. Determine: a) the thermal efficiency of the cycle (%) and b) the net power output of the power plant (kW).

Answers

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

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

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]

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