An object at a vertical elevation of 20 m and a speed of 5 m/s decreases in elevation to an elevation of 1 m. At this location, the object has a velocity of 15 m/s. The mass of the object is 68 kg. Assuming the object is the system, determine if there is any work transfer associated with the object (there is no heat transfer). The object is solid, incompressible and its temperature does not change during the process. If there is work transfer, is work done on or by the object? Assume the acceleration of gravity g = 9.81 m/s2.

Answer:

The velocity of the pulling plate is 14.98 m/s.

Explanation:

The shear stress caused by the 20 Newton force on the plate is given by

[tex]\tau =\frac{20}{Area}=\frac{20}{0.3\times 1.0}=66.66N/m^2[/tex]

Now by newton's law of viscosity we have

[tex]\tau =\mu\cdot \frac{\Delta u}{\Delta y}[/tex]

where

[tex]\mu [/tex] is the dynamic viscosity of the water at 20 degree

[tex]\frac{\Delta u}{\Delta y}[/tex] is the velocity gradient

Applying values we get

[tex]66.66=8.90\times 10^{-4}\times \frac{U}{0.2\times 10^-3}\\\\\therefore u=14.98m/s[/tex]

The CM firm's responsibilities under a construction management contract should begin at the planning phase. This early engagement allows for a cohesive approach to project management, involving definition of scope, budgeting, and risk assessment. Construction managers work closely with designers and contractors throughout the project, ensuring it meets design specifications and safety standards.

Under a construction management (CM) contract, the CM firm's responsibilities should begin at the planning phase, which precedes the design phase, construction phase, and conceptual phase. During the planning phase, the CM firm is involved in defining project objectives, scope, and feasibility, providing valuable input that shapes the overall project. The expertise of construction managers is utilized to help with site selection, project scheduling, budgeting, and risk management. Their early involvement ensures that the construction phase flows more smoothly, as potential issues may have been identified and addressed beforehand. Construction managers play a crucial role in maintaining the integrity of the design and adherence to specifications throughout the construction process.

Designers, on the other hand, continue their involvement even after generating detailed technical drawings and digital models by acting as on-site supervisors during construction. Their responsibility remains to ensure that contractors follow the design's specifications, indicating direct involvement in the contract construction phase. The construction site can present various risks, including the exposure of workers to dangerous materials which designers need to consider while specifying project materials. The collaborative effort between the CM firm, designers, and contractors is fundamental to achieving a successful and safe project completion, fulfilling the client's expectations, and managing any equity issues related to contract allocations.

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

[tex]\Delta x=\frac{PL^3}{3EI}[/tex]

Hence we can write

[tex]P=\frac{3EI\cdot \Delta x}{L^3}[/tex]

Comparing with the standard spring equation [tex]F=kx[/tex] we find the cantilever analogous to spring with [tex]k=\frac{3EI}{L^3}[/tex]

Now the angular frequency of a spring is given by

[tex]\omega =\sqrt{\frac{k}{m}}[/tex]

where

'm' is the mass of the load

Thus applying values we get

[tex]\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}[/tex]

[tex]\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec[/tex]

Answer:

its a lil confusing

Explanation:

Answer:

105mg of calcium chloride [tex]CaCl_{2}[/tex]

Explanation:

The molecular formula of calcium chloride is [tex]CaCl_{2}[/tex] and the molecular formula of soda ash is [tex]Na_{2}CO_{3}[/tex]

First of all you should write the balanced reaction between both compounds, so:

[tex]CaCl_{2}+Na_{2}CO_{3}=2NaCl+CaCO_{3}[/tex]

Then you should have the molar mass of the two compounds:

Molar mass of [tex]CaCl_{2}=110.98\frac{g}{mol}[/tex]

Molar mass of [tex]Na_{2}CO_{3}=105.98\frac{g}{mol}[/tex]

Now with stoichiometry you can find the mass of calcium chloride that reacts with 100mg of soda ash, so:

[tex]100mgNa_{2}CO_{3}*\frac{1molNa_{2}CO_{3}}{105.98gNa_{2}CO_{3}}*\frac{1molCaCl_{2}}{1molNa_{2}CO_{3}}*\frac{110.98gCaCl_{2}}{1molCaCl_{2}}=105mgCaCl_{2}[/tex]

Answer:

From -2147483647 to 2147483648.

Explanation:

A 32 bit integer has 2^32 = 4294967296 possible values. A signed integer has positive and negative values as well as the zero.

Of these values (2^32)/2 will be positive, (2^32)/2 - 1 will be negative and one will be the zero.

Therefore the range is from -(2^32)/2 + 1 to (2^32)/2.

This can also be expressed as from -2147483647 to 2147483648.

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid  F

[tex]F=\rho AV^2[/tex]

So force exerted in x-direction

[tex]F_x=\rho AV^2[/tex]

[tex]F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2[/tex]

F=0.763 N

So force exerted in y-direction

[tex]F_y=\rho AV^2[/tex]

[tex]F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2[/tex]

F=0.763 N

So the resultant force R

[tex]R=\sqrt{F_x^2+F_y^2}[/tex]

[tex]R=\sqrt{0.763^2+0.763^2}[/tex]

R=1.079

So the force required to hold the pipe is 1.08 N.

Answer:

1.29 lbf

Explanation:

Weight is a force, it is the product of a mass by the acceleration of gravity.

f = m * a

Gallons are a unit of volume. The relationship of volume with mass is:

m = V * δ

δ is the density (mass per unit of volume)

Then the weight of a liquid is:

w = V * δ * g

Rearranging:

δ = w / (V * g)

The acceleration of gravity on Earth is 32.2 ft/s^2

δ = 50 / (20 * 32.2) = 0.078 lbm/gallon

Knowing this and the gravity on the Moon we can calculate how much would 3 gallons of this liquid weight on the Moon.

w = V * δ * g

w = 3 * 0.078 * 5.51 = 1.29 lbf

Answer:

The given statement is false.

Explanation:

The basic equation of motion for a control volume is as follows

[tex]\frac{d\overrightarrow{p}}{dt}=\int_{c.v}\frac{\partial }{\partial t}(\rho v)dV+\int_{cs}(\rho \overrightarrow{v})\cdot \overrightarrow{v_{r}}.\widehat{n}dS[/tex]

the symbols have the usual meaning as

[tex]\rho [/tex] is density of the fluid

[tex]v [/tex] is the velocity of the fluid

[tex]\widehat{n}[/tex] is the direction vector of area over which the integration is carried out

As we see that the terms in the right hand side of the equation is not zero if the flow is unsteady or the velocity is changing in the control volume the term in the left is non zero hence the momentum is not conserved.

Answer:4050 W

Explanation:

Given

Heat transfer Coefficient(h)=[tex]20 W/m^2-K[/tex]

Air temperature =75 F

surface area(A)=[tex]7.5 m^2[/tex]

Temperature of hot tube is 102 F

We know heat transfer due to convection is given by

[tex]Q=hA\left ( \Delta T\right )[/tex]

[tex]Q=20\times 7.5\left ( 102-75\right )=4050 W[/tex]

Answer: 0.57 hp,  24.2 Btu/min

Explanation:

Hi!

The power P delivered when rotating a crank at angular speed ω, applying a torque τ is given by:

P = τ*ω

In this case torque is τ = 12 in * 25 lbf = 300 lbf*in

Angular speed is 2 round per seconds, which is 120 RPM. Then power is:

P = 300*120 lbf*in*RPM

1 hp = (lbf*in*RPM) / 63,025

P = 0.57 hp

1 hp = 42,4 Btu/min

Then P = 24.2 Btu/min

Answer:

The three major network categories are LAN (Local Area Network), MAN (Metropolitan Area Network) and WAN (Wide Area Network)

Explanation:

A Local Area Network (LAN) is the simplest form of a network. It is usually limited to a geographical area and consists of a group of computers in the same organisation linked together. In most cases, the same technology is used for all computers in this network (eg. ethernet).

A Metropolitan Area Network (MAN) can connect multiple LANs as long as they are close to each other (geographically). MANs allow high speed remote connection as if the devices were on the same LAN.

A Wide Area Network (WAN) can connect multiple LANs across large geographical distances. Unlike MANs, the speed might differ based on factors such as distance. The internet is the most popular example of a WAN.

Answer:

 Load =  2.42 tons

Lifting speed = 24 RPM

Explanation:

Given that

Power=10 KW

Efficiency = 90%

So the actual power,P=0.9 x 10 =9 KW

Speed of motor = 1440 RPM

Diameter of drum = 300 mm

radius =150 mm

G=60:1

Lets take speed of drum =N

We know that

[tex]G=\dfrac{Speed\ of\ motor}{Speed\ of\ drum}[/tex]

[tex]60=\dfrac{1440}{N}[/tex]

N=24 RPM

We know that

[tex]P=\dfrac{2\pi NT}{60}[/tex]

Where T is the torque

[tex]9000=\dfrac{2\pi \times 24\times T}{60}[/tex]

T=3580.98 N.m

Lets take load =F

So T= F x r

3580.98 = F x 0.15

F=23.87 KN

We know that

1 KN=0.109 tons

So 23.87 KN= 2.42 tons

So the load =  2.42 tons

Lifting speed = 24 RPM

Answer:

[tex]Q_d = 0.0166 J/mol[/tex]

Explanation:

given data:

diffusion coefficient [tex]= 3.98 \times 10^{-13} m^2/s[/tex]

[tex]T_1 = 980 Degree\ celcius = 1253 K[/tex]

[tex]T_2 = 650 Degree\ celcius = 923 K[/tex]

[tex]D_2 = 3.55\times 10^{-16} m^2/s[/tex]

Activation energy is given as[tex] = -2.3R \frac{ \Delta log D}{\Delta\frac{1}{T}}[/tex]

[tex] Q_d = -2.3 R [\frac{logD_1 - logD_2}{ \frac{1}{T_1} - \frac{1}{T_2}}][/tex]

[tex]Q_d = -2.3 \times 8.31  \frac{log(3.98*10^{-13} - log(3.55*10^{-16}}{ \frac{1}{1253} - \frac{1}{923}}[/tex]

[tex]Q_d = 0.0166 J/mol[/tex]

Answer:

q = (709*K*i + 0*j + 0*k)

Explanation:

1/4 inch is 6.35 mm = 0.00635 m

A = 0.9 * 1.65 = 1.48 m^2

t1 = 1 F = -17.2 C

t2 = -7 F = -21.7 C

The heat will be conducted through the glass. We use a frame of reference with the origin on the warmer side of the glass (t1) and the positive X axis pointing perpendicular to the glass towards the colder side.

The Fourier equation for plates (one dimensional form) is:

q = -K * dT / dx

We can simplify  dT/dx to ΔT/Δx if we assume the temperature changes linearly through the glass.

q = -K * ΔT/Δx

K is the thermal conductivity of the glass

ΔT = t2 - t1

ΔT = -21.7 + 17.2 = -4.5 C

Δx is the thickness of the glass

q = -K * -4.5 / 0.00635

q = 709 * k

The vector is: q = (709*K*i + 0*j + 0*k)

It has no components in Y or Z because there if no variation of temperature in those directions (the temperature of the glass is given as each face having a constant temperature throughout its area).

Answer:

23.2 m/s

Explanation:

A jetpack appliang a force on a body would result in an acceleration following Newton's equation:

f = m * a

Rearranging:

a = f / m

The problem doesn't state a mass, so I'll use mine, which is

m = 72 kg

Then:

a = 334 / 72 = 4.64 m/s^2

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

Since I was standing before turning the jetpack on

V0 = 0

So:

V(5) = 0 + 4.64 * 5 = 23.2 m/s

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

[tex] P = T\times \omega[/tex]

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

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

Therefore Power is

[tex]P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt[/tex]

P = 80.922 KW

Answer:

Power output, [tex]P_{out} = 178.56 kW[/tex]

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, [tex]T = 350^{\circ}C[/tex]

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, [tex]\dot{m} = 0.1 kg.s^{- 1}[/tex]

Diameter of exhaust pipe, [tex]d_{h} = 15 cm = 0.15 m[/tex]

Pressure at exhaust, P' = 50 kPa

temperature, T' =  [tex]100^{\circ}C[/tex]

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

[tex]\dot{m} = \frac{Av_{i}}{V_{1}}[/tex]

[tex]0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}[/tex]

[tex]0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}[/tex]

[tex]v_{i} = 3.986 m/s[/tex]

Now, eqn for compressible fluid:

[tex]\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}[/tex]

Now,

[tex]\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}[/tex]

[tex]\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}[/tex]

[tex]\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}[/tex]

[tex]v_{e} = 19.33 m/s[/tex]

Now, the power output can be calculated from the energy balance eqn:

[tex]P_{out} = -\dot{m}W_{s}[/tex]

[tex]P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}[/tex]

[tex]P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW[/tex]

The power output of the Turbine is; 225.69 kW

We are given

Pressure of steam; P = 1400 kPa

Temperature of steam at state 1; T = 350°C

Diameter of pipe; d₁ = 8 cm = 0.08 m

Mass flow rate; m' =  0.1 kg/s

Diameter of exhaust pipe; d₂ = 15 cm = 0.15 m

Pressure at exhaust; P' = 50 kPa

Temperature at state 2; T' =  100°C

Area; A = πd²/4

A = π * 0.08²/4

A = 0.0016π m²

We can find the find initial velocity from the formula;

v₁ = m' * V₁/A

v₁ = (0.1 * 0.2004/(0.0016π))

v₁ = 3.986 m/s

From equation of compressible fluid, we know that;

(A₁ * v₁)/V₁ = (A₂ * v₂)/V₂

A₂ = πd₂²/4

A₂ = π * 0.15²/4

A₂ = 0.005625π m²

(0.0016π * 3.986)/0.2004  = (0.005625π * v₂)/3.4181

Solving for v₂ gives;

v₂ = 19.33 m/s

Finally power output is gotten from  the expression;

P_out = -m'(H₂ - H₁) + (v₂² - v₁²)/2

P_out = -0.1(2682.6 - 3150.7) + (19.33² - 3.986²)/2

P_out = 225.69 kW

Read more about Power Output at; https://brainly.com/question/25573309

Answer:

Maximum and mean velocity will be equal i.e 0.5 m/s

Explanation:

given data:

viscosity = 0.001 m^2/s

Thickness of thin film = 2 mm = 0.002 m

Neglecting the body weight hence no shear stress

Mean velocity is given V

[tex]V = \frac{\nu}{D}[/tex]

[tex]V = \frac{0.001}{2*10^{-3}}[/tex]

V = 0.5 m/s

Maximum and mean velocity will be equal i.e 0.5 m/s

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

[tex]Losses=K\dfrac{V^2}{2g}[/tex]

Answer with Explanation:

The crown will be pure if it's specific gravity is 19.3

Now by definition of specific gravity it is the ratio between the weight of an object to the weight of water of equal volume

Since it is given that the weight of the crown is 11.8 N we need to find it's volume

Now According to Archimedes principle when the crown is immersed into water the water shall exert a force in upwards direction on the crown with a magnitude equaling to weight of the water displaced by the crown

Mathematically this is the difference between the weight of the crown in air and weight when immersed in water

Thus Buoyant force is [tex]F_{B}=11.8-10.9=0.9N[/tex]

Now by Archimedes principle This force equals in magnitude to the weight of water of same volume as of the crown

Thus the specific gravity of the crown equals

[tex]S.G=\frac{11.8}{0.9}=13.11[/tex]

As we see that the specific gravity of the crown material is less than that of pure gold hence we conclude that it is impure.  

Given:

[tex]\to SG_{gold} = 19.3 \\\\[/tex]

weight of air [tex]W_{air} = 11.8\ N \\\\[/tex]

water weight [tex]W_{water} = 10.9 \ N \\\\[/tex]

To Find:

using the buoyancy B that calculates the difference of weight=?

Solution:

Using formula:

[tex]\to B = W_{air} - W_{water} \\\\[/tex]

       [tex]= 11.8\ N - 10.9\ N\\\\ = 0.9 \ N\\\\[/tex]

Using a formula for calculating weight:

[tex]\to W_{air} = SG_{\gamma water}\times V_{crown}\\\\ \to W_{water} = B(SG - 1)[/tex]

calculating the [tex]\bold{SG_{crown}}[/tex]:

[tex]\to SG_{crown}=1+\frac{W_{water}}{B} \\\\[/tex]

                  [tex]= 1 + \frac{10.9\ N}{0.9\ N}\\\\= 1 + 12.11\\\\= 13.11[/tex]

Knowing that the [tex]SG_{gold}= 19.3[/tex] indicates that the crown is not made of pure gold.

[tex]\to \bold{SG_{crown}= 13.1}[/tex] The crown is not composed entirely of gold.

Find out more information about the gold here:

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Answer:

[tex]\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right][/tex]

Explanation:

A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of  the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.

If there is any question, just let me know.

Answer:

False

Explanation:

Vernier caliper is used for measuring very small distances. It is very precise instrument for measuring very small distances.

With the help of vernier caliper we can measure distances from 6 inch to about 12 inch

But in question we have given 0.001 in so it is impossible to measure 0.001 inch from vernier caliper so it is a false statement

Answer:

The pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.

Explanation:

Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 160°F.  

Given:  

Mass of the water is 10 lb.

Dryness fraction is 0.5.

Temperature of water is 160°F.  

From steam table at 160°F:  

The pressure in the tank is 4.74703 psi.  

Specific volume of saturated water is 0.0163918 ft³/lb.  

Specific volume of saturated steam is 77.1773 ft³/lb.  

Calculation:  

Step1  

From steam table at 160°F:  

The pressure in the tank is 4.74703 psi.  

Step2  

Specific volume of tank is calculated as follows:  

[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]

[tex]v=0.0163918 +0.5(77.1773 -0.0163918)[/tex]

[tex]v=0.0163918 +38.58045[/tex]

v=38.5968 ft³/lb.  

Step4  

Volume is calculated as follows:  

[tex]V=v\times m_{t}[/tex]

[tex]V=38.5968\times10[/tex]

V=385.968 ft³.  

Thus, the pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.

Answer:

Star topology

Explanation:

Topology:

Topology is the connection of networks .These network connect by nodes.

Type of topology:

1.Bus topology

2.Star topology

3.Ring topology

4.Mesh topology

Star topology is the most reliable topology .Because failure of node node does not affect the other nodes.In star topology  network  are connect in star form.

Ring topology is the most undesirable topology.In ring topology network connects in the ring form.In ring topology failure of one node affect the whole network.

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, [tex]T_{i} = 20^{\circ}C[/tex] = 273 + 20 = 293 K

Temperature at the outlet, [tex]T_{o} = 200{\circ}C[/tex] = 273 + 200 = 473 K

Pressure at inlet, [tex]P_{i} = 80 kPa = 80\times 10^{3} Pa[/tex]

Pressure at outlet, [tex]P_{o} = 800 kPa = 800\times 10^{3} Pa[/tex]

Speed at the outlet, [tex]v_{o} = 20 m/s[/tex]

Diameter of the tube, [tex]D = 10 cm = 10\times 10^{- 2} m = 0.1 m[/tex]

Input power, [tex]P_{i} = 400 kW = 400\times 10^{3} W[/tex]

Now,

To calculate the heat transfer, [tex]Q[/tex], we make use of the steady flow eqn:

[tex]h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}[/tex]

where

[tex]h_{i}[/tex] = specific enthalpy at inlet

[tex]h_{o}[/tex] = specific enthalpy at outlet

[tex]v_{i}[/tex] = air speed at inlet

[tex]p_{s}[/tex] = specific power input

H and H' = Elevation of inlet and outlet

Now, if

[tex]v_{i} = 0[/tex] and H = H'

Then the above eqn reduces to:

[tex]h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}[/tex]

[tex]Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}[/tex]                (1)

Also,

[tex]p_{s} = \frac{P_{i}}{ mass, m}[/tex]

Area of cross-section, A = [tex]\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}[/tex]

Specific Volume at outlet, [tex]V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s[/tex]

From the eqn:

[tex]P_{o}V_{o} = mRT_{o}[/tex]

[tex]m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s[/tex]

Now,

[tex]p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg[/tex]

Also,

[tex]\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg[/tex]

Now, using these values in eqn (1):

[tex]Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW[/tex]

Now, rate of heat transfer, q:

q = mQ = [tex]0.925\times 813.33 = 752.33 kW[/tex]

Answer:

The shear plane angle is 34.45°.

Explanation:

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the edge of the surface. For orthogonal cutting operation feed is the chip thickness.

Given:

Rake angle is 15°.

Uncut chip thickness is 0.5 mm.

Speed is 55 m/min.

Feed or the chip thickness is 0.3 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

[tex]r=\frac{t_{c}}{t}[/tex]

[tex]r=\frac{0.3}{0.5}[/tex]

r = 0.6

Step2

Shear angle is calculated as follows:

[tex]tan\phi=\frac{rcos\alpha}{1-rsin\alpha}[/tex]

Here, [tex]\phi[/tex] is shear plane angle, r is chip reduction ratio and [tex]\alpha[/tex] is rake angle.

Substitute all the values in the above equation as follows:

[tex]tan\phi=\frac{rcos\alpha}{1-rsin\alpha}[/tex]

[tex]tan\phi=\frac{0.6cos15^{\circ}}{1-0.6sin15^{\circ}}[/tex]

[tex]tan\phi=\frac{0.57955}{0.8447}[/tex]

[tex]\phi=34.45^{\circ}[/tex]

Thus, the shear plane angle is 34.45°.

Explanation:

Styrene is  a vinyl monomer in which there is a carbon carbon double bond.

The polymerization of the styrene, which is initiated by using a free radical which reacts with the styrene and the compound thus forms react again and again to form polystyrene (PS).

The equation is shown below as:

[tex]\begin{matrix}& C_6H_5 \\&|\\n H_2C & =CH\end{matrix}[/tex]        ⇒                [tex]\begin{matrix}&C_6H_5 \\&|\\ -[-H_2C & -CH-]-_n\end{matrix}[/tex]

Answer:

Maximum weight = 76.64 lb

Explanation:

In the second figure attached it can be seen the free body diagram  of the problem.

The equation of equilibrium respect x-axis is

[tex] \sum F_x = 0 [/tex]

[tex] F_{AC} \times sin 30 - F_{AB} \times \frac{3}{5} = 0 [/tex]

[tex] F_{AC} = 1.2 \times F_{AB} [/tex]

So, cable segmet AC supports bigger tension than segment AB. If [tex] F_{AC} = 50 lb [/tex] then

[tex] F_{AB} = \frac{F_{AC}}{1.2} [/tex]

[tex] F_{AB} = 41.67 lb [/tex]

The equation of equilibrium respect y-axis is

[tex] \sum F_y = 0 [/tex]

[tex] F_{AB} \times \frac{4}{5} + F_{AC} \times cos 30 - W = 0 [/tex]

[tex] 41.67 lb \times \frac{4}{5} + 50 lb \times cos 30 - W = 0 [/tex]

[tex] W = 76.64 lb [/tex]

Answer:

0.49%

Explanation:

Assuming this is about steel. Higher carbon steel tend to be harder and stronger, but that hardness makes them more brittle and less ductile. This is because with higher carbon concentrations there is more formation of pearlite, which has higher hardness than ferrite. Pearlite contains layers of cementite, which is a fragile material, so the presence of these microstructures will make the whole thing more brittle.

Answer:

The pressure is 37.56 atm.

Temperature at the end of adiabatic compression is 825.73 K.

Explanation:

Given that

[tex]T_1=20C[/tex]

[tex]P_1=1\ atm[/tex]

[tex]V_1=800\ cm^3[/tex]

[tex]V_2=60\ cm^3[/tex]

So r=800/60=13.33

γ=1.4

For process 1-2

[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]

[tex]\dfrac{T_2}{273+20}=13.33^{1.4-1}[/tex]

[tex]T_2=825.73\ K[/tex]

[tex]\dfrac{P_2}{P_1}=r^\gamma[/tex]

[tex]\dfrac{P_2}{1}=13.33^{1.4}[/tex]

[tex]P_2=37.56\ atm[/tex]

So

The pressure is 37.56 atm.

Temperature at the end of adiabatic compression is 825.73 K.