A substance temperature was 62 deg R. What is the temperature in deg C? A.) 50.7 B.) 45.54 C) 80.0 D) 94.4

Answer:

The length of bar will be 2.82 m

Explanation:

Given that

d= 15 mm

r= 7.5 mm

Shear stress = 110 MPa

θ =  30°                                  (30°   = 30°  x π/180°  =0.523 rad)

θ = 0.523 rad

G for steel

G= 79.3 GPa

We know that

[tex]\dfrac{\tau}{r}=\dfrac{G\theta }{L}[/tex]

[tex]\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}[/tex]

L= 2. 82 m

The length of bar will be 2.82 m

The angular speed of a flywheel rotating at 300 revolutions per minute is calculated by converting revolutions to radians and minutes to seconds, resulting in an angular speed of 10π rad/s.

The question asks to determine the angular rate of rotation of a searchlight on a boat when targeting an automobile that is 3000 ft away and moving at 80 ft/s. This involves understanding the relationship between linear velocity, radius, and angular velocity, which is a crucial concept in trigonometry and physics.

To find the angular speed of the flywheel rotating at 300 revolutions per minute, we first convert revolutions per minute (rpm) to radians per second, knowing that one revolution is 2π radians and there are 60 seconds in a minute. Therefore:

Angular speed = 300 rpm × (2π radians/revolution) × (1 minute/60 seconds) = 300 × 2π / 60 rad/s = 10π rad/s.

Answer:29,930 kJ

Explanation:

Given

Car starts with an initial elevation of 500 m and drives up a hill to reach a final elevation of 2000 m

Final velocity (V)=20 m/s

Energy of car increases by 100 kJ

mass of car(m)=2000 kg

[tex]Total Energy =\Delta PE+\Delta KE+\Delta U[/tex]

[tex]\Delta PE=mg(\Delta h)=2000\times 9.81\times (2000-500)[/tex]

[tex]\Delta PE=29,430 kJ[/tex]

[tex]\Delta KE=m\frac{v_2^2-v_1^2}{2}[/tex]

[tex]\Delta KE=2000\times \frac{20^2-0^2}{2}[/tex]

[tex]\Delta KE=400 kJ[/tex]

[tex]\Delta U=100 kJ[/tex]

Total Energy=29,430+400+100=29,930 kJ

Answer:

The diameter of the test cylinder should be 7.65 meters.

Explanation:

The Hooke's law relation between stress and strain is mathematically represented as

[tex]Stress=E\times strain\\\\\sigma =e\times \epsilon[/tex]

Where 'E' is modulus of elasticity of the material

Now by definition of strain we have

[tex]\epsilon =\frac{\Delta L}{L_{o}}[/tex]

Applying values to obtain strain we get

[tex]\epsilon =\frac{0.5}{380}=0.001316[/tex]

Thus the stress developed in the material equals

[tex]\sigma = 110\times 10^{3}\times 0.001316=144.76N/m^{2}[/tex]

Now by definition of stress we have

[tex]\sigma =\frac{Force}{Area}\\\\\therefore Area=\frac{Force}{\sigma }\\\\\frac{\pi D^{2}}{4}=\frac{6660N}{144.76}=46m^{2}[/tex]

Solving for 'D' we get

[tex]D=\sqrt{\frac{4\times 46}{\pi }}=7.653meters[/tex]

Answer:

Point to point control motion system:

  In point to point control motion system tool perform specific task at a particular location.Point to point control motion system is also called positioning system.It perform intermittent operation.

Ex:  Drilling operation is a point to point motion control system.

Continuous path control system:

 Continuous path control system is continuous operation that perform by tool.The program used in  continuous path control system is more complex than point to point motion control system.

Ex :Milling operation is a  continuous path control system.

Answer:

W=-280.67 KJ

Explanation:

Given that

Initial pressure = 150 KPa

Final pressure = 600 KPa

Temperature T= 285 K

Mass m=2.4 Kg

We know that ,work in isothermal process given as

[tex]W=mRT\ ln\dfrac{P_1}{P_2}[/tex]

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

[tex]W=mRT\ ln\dfrac{P_1}{P_2}[/tex]

[tex]W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}[/tex]

W=-280.67 KJ

Negative sign indicates that it is compression process and work is done on the gas.

To avoid cavitation for water at 160 °F flowing through a sharp bend, the minimum absolute pressure should be slightly above the vapor pressure of water at this temperature, which is approximately 0.363 psi.

When water flows through a sharp bend, cavitation can occur if the local pressure falls to or below the fluid's vapor pressure. To estimate the minimum absolute pressure without causing cavitation for water at 160 °F, we must consider water's vapor pressure at this temperature. At 160 °F (about 71 °C), the vapor pressure of water is approximately 0.363 psi. Since fluids cannot have a negative absolute pressure, and to avoid cavitation, the absolute pressure must stay above this vapor pressure. Therefore, considering atmospheric pressure to be approximately 14.7 psi, to avoid cavitation, the minimum absolute pressure in the system should be slightly above 0.363 psi to ensure no cavitation occurs.

Answer:

TRUE

Explanation:

Polymers can be natural as well as synthetic

The polymer which are found in nature are called natural polymer tease polymer are not synthesized, they are found in nature

Example of natural polymers is cellulose, proteins etc

On the other hand synthetic polymers are not found in nature they are synthesized in market

There are many example of synthetic polymer

Example : nylon, Teflon etc  

So it is a true statement

Answer:

2% to 20% Ni

Explanation:

If we will talk about steel then ,for increasing the toughness property  of steel generally 2% to 20% Ni added  .Ni also increase resistance to corrosion and oxidation, impact strength and strength.

We know that steel is an alloy of iron and carbon.But to improve the property of steel different alloying elements added by this steel become desirable to use at different situations.

Answer:

water=14.1 kg

air=0.1348kg

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

To solve this exercise we must find the specific volume of water and air in the two states and then subtract them and multiply them by the volume of the tank to find the change in mass, the following equation is inferred

Δm=V(ρ2-ρ1)

Where V= tank volume=500L=0.5m^3

ρ2= density of fluid in state 2

ρ1=density of fluid in state 1

Δm=change of mass

for water

ρ1=971.8kg/m^3(80C)

ρ2=1000kg/m^3(5C)

Δm=0.5(1000-971.8)

Δm=14.1 kg

for air

ρ1=0.9994kg/m^3(80C)

ρ2=1.269kg/m^3(5C)

Δm=0.5(1.269-0.9994)

Δm=0.1348kg

The two basic network administration models are centralized and decentralized, each with distinct advantages and challenges that affect network performance, security, and scalability.

The two basic network administration models are centralized and decentralized. In a centralized model, network control and decision-making are located at a single point, typically within a dedicated device or group of servers. Conversely, a decentralized model distributes control across multiple nodes, allowing for individual nodes to operate independently while still being part of the overall network.

Understanding these models is crucial for designing efficient networks that cater to specific organizational needs and for implementing dynamics on network models, such as discrete state/time models or continuous state/time models. Each model presents different advantages and challenges that can influence network performance, security, and scalability.

Answer:

Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.

Explanation:

Step1

Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.  

Step2

All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.  

Answer:

Hybrid topology is the connection of one or more than one topology.

Explanation:

Topology:

 Topology is the arrangement of network.These network connects by line and nodes.

Type of topology:

1.Bus topology

2.Star topology

3.Ring topology

4.Mesh topology

Along with given above topology one topology is also used is known as hybrid topology.Hybrid topology is the connection of one or more than two one above given topology.

Answer:

The pressure inside the tire is [tex]0.304\frac{N}{mm^{2}}[/tex]

Explanation:

The pressure gauge indicates the difference between the atmospheric pressure and the pressure inside the tire, so we have the following equation:

Pressure inside the tire = Gauge pressure + Atmospheric pressure

Where the gauge pressure is given in the problem and is 29.35psi and the atmospheric pressure is 14.7psi.

Replacing the values, we have:

Pressure inside the tire = 29.35psi + 14.7psi

Pressure inside the tire =  44.05psi

Now we have to convert from psi to [tex]\frac{N}{mm^{2}}[/tex], so:

44.05psi = [tex]44.05\frac{lbf}{in^{2}}[/tex]

[tex]44.05\frac{lbf}{in^{2}}*(\frac{1in}{25.4mm})^{2}*\frac{4.4482N}{lbf}=0.304\frac{N}{mm^{2}}[/tex]

Answer:

origin

Explanation:

The point of intersection of axis is called origin.

In 2D origin is the intersection point of x-axis and y-axis if we go right to the origin then it is positive x axis, if we go left side of origin then it is negative x- axis

Similarly when we go above the origin then it positive y axis , and if we go bellow the origin then it is negative x axis

In 3D origin is the intersection of x-axis, y-axis and z-axis

NOTE- For defining i take here x axis as horizontal axis and y-axis as vertical axis

Answer:

The maximum length of the specimen is 0.2927 m or 292.7 mm

Solution:

Modulus of elasticity, E = 126 GPa = [tex]126\times 10^{9}[/tex]

Diameter of the cross-section, D = 4.0 mm = [tex]4.0\times 10^{- 3} m[/tex]

Force due to tension, F = 2380 N

Maximum elongation, [tex]\Delta L = 0.44 mm = 0.44\times 10^{- 3} m[/tex]

Now,

The maximum length of the specimen, [tex]L_{m}[/tex] can be calculated as follows:

The cross-sectional area, [tex]A_{c} = \frac{\pi D^{2}}{4} = \frac{\pi\times (4.0\times 10^{- 3})^{2}}{4} = 1.256\times 10^{- 5} m^{2}[/tex]

Now, the stress on the specimen, [tex]\sigma_{s} = \frac{F}{A_{c}} = \frac{2380}{1.256+\times 10^{- 5}}[/tex]

[tex]\sigma_{s} = 1.89\times 10^{8} N/m^{2}[/tex]

Now,

The strain on the specimen, [tex]\epsilon_{s}[/tex]:

[tex]\epsilon_{s} = \frac{\Delta L}{L_{m}}[/tex]

Also, from Hooke's law:

[tex]E = \frac{\sigma_{s}}{epsilon_{s}}[/tex]

⇒ [tex]E = \frac{1.89\times 10^{8}}{\frac{\Delta L}{L_{m}}}[/tex]

⇒ [tex]L_{m} = \frac{\Delta Ltimes E}{1.89\times 10^{8}}[/tex]

⇒ [tex]L_{m} = \frac{0.44\times 10^{- 3}\times 126\times 10^{9}}{1.89\times 10^{8}} = 0.2927 m[/tex]

The maximum length of the specimen before deformation is:           292.72 mm (0.2927 m).

For solving this question, it's necessary to know some concepts about the material's properties.

The tensile stress (σ) is determined from the ratio between load and original area before the load applied (σ=[tex]\frac{F}{Ao}[/tex]). Depending on the load applied, the material can have an elastic deformation (temporary deformation) and plastic deformation (permanent deformation). Both deformations can be calculated by the equation: ε=ΔL/Lo, where ΔL= deformation elongation and Lo= the original length before the load applied.

When the material is in the elastic portion, there is a linear relationship between stress and strain given by: σ=Eε. Due to this relationship, it is possible to find the elastic deformation (ε) when we know the stress (σ) and elastic modulus (E).

Now you have the necessary information to solve your question.

The question gives:

E (elastic modulus) =126 GPA

d (original cross-sectional diameter)=4 mm

F (tensile load)=2380 N

ΔL (maximum allowable elongation) =0.44 mm

       1. Find the area of the cylindrical specimen.

       2. Find the tensile stress.

          σ= [tex]\frac{F}{Ao} =\frac{2380 N }{12.57 mm^2} =189.39 MPa[/tex]

      3. Calculate the maximum length of the specimen before deformation.

       Knowing that  ε=ΔL/Lo and σ=Eε, you can rewrite these equations as:

                                      σ= E * (ΔL/Lo)

                                      σ= (E * ΔL)/Lo

   The question asks the maximum length of the specimen before deformation, therefore you should find Lo. Thus,

                                     Lo= (E * ΔL)/σ

                  [tex]Lo=\frac{126*10^3 MPa*0.44 mm}{189.39 MPa} =292.72 mm= 0.2927 m[/tex]

Read more about the tensile stress here:

https://brainly.com/question/19756298

Answer with Explanation:

By the equation or Torque we have

[tex]\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}[/tex]

where

T is the torque applied on the shaft

[tex]I_{p}[/tex] is the polar moment of inertia of the shaft

[tex]\tau [/tex] is the shear stress developed at a distance 'r' from the center of the shaft

[tex]\theta [/tex] is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft [tex]I_{p}=\frac{\pi R^4}{2}[/tex]

For a hollow shaft [tex]I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}[/tex]

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

[tex]\frac{2T}{\pi R^4}\times r=\tau _{solid} [/tex]

2) For hollow shaft we have

[tex]\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4} [/tex]

Comparing the above 2 relations we see

[tex]\frac{\tau _{solid}}{\tau _{hollow}}=0.76[/tex]

Similarly for angle of twist we can see

[tex]\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316[/tex]

Part b)

Strength of solid shaft = [tex]\tau _{max}=\frac{T\times R}{I_{solid}}[/tex]

Weight of solid shaft =[tex]\rho \times \pi R^2\times L[/tex]

Strength per unit weight of solid shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}[/tex]

Strength of hollow shaft = [tex]\tau '_{max}=\frac{T\times R}{I_{hollow}}[/tex]

Weight of hollow shaft =[tex]\rho \times \pi (R^2-0.7R^2)\times L[/tex]

Strength per unit weight of hollow shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}[/tex]

Thus [tex]\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16[/tex]

Answer:

W=-940.36 KJ

Explanation:

Given that

[tex]P_1=1\ bar,V_1=4.25 {m^3}[/tex]

[tex]P_2=6.8\ bar[/tex]

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

Now by putting the values                (1.4 bar =140 KPa)

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=140\times 4.25 \ln \dfrac{1.4}{6.8}[/tex]

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

[tex]60.1cm*\frac{1m}{100cm}=0.601m[/tex]

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

[tex]V_{water}=\pi  r^{2}h[/tex]

[tex]V_{water}=\pi (\frac{0.601m}{2})^{2}*120m[/tex]

[tex]V_{water}=113.28m^{3}[/tex]

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

[tex]d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}[/tex]

[tex]d_{water}=1000\frac{Kg}{m^{3}}[/tex]

Now, water density is given by the equation [tex]d=\frac{m}{V}[/tex], where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

[tex]m_{water}=d_{water}.V_{water}[/tex]

[tex]m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}[/tex]

[tex]m_{water}=113280Kg[/tex]

With the water mass we can find the weight of water:

[tex]w_{water}=m_{water} *g[/tex]

[tex]w_{water}=113280kg*9.8\frac{m}{s^{2}}[/tex]

[tex]w_{water}=1110144N[/tex]

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

[tex]w_{total}=w_{water}+w_{pipe}[/tex]

[tex]w_{total}=1110144N+2500N[/tex]

[tex]w_{total}=1112644N[/tex]

Converting this total weight to kN, we have:

[tex]1112644N*\frac{0.001kN}{1N}=1113kN[/tex]

Answer:

-273.16 °C

-459.677 °F

0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= [tex]2\pi\sqrt{\frac{L}{g}}[/tex]

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= [tex]2\pi\sqrt{\frac{L}{9.81}}[/tex]

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = [tex]\frac{\textup{AE}}{\textup{L}}[/tex]

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = [tex]\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}[/tex]

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = [tex]\sqrt{\frac{K}{m}}[/tex]

or

mass, m = [tex]\sqrt{\frac{K}{\omega_n^2}}[/tex]

or

mass, m = [tex]\sqrt{\frac{20.66\times10^8}{0.95^2}}[/tex]

mass, m = 22.90 × 10⁸ kg

Explanation:

Balancing:

  Generally balancing are of two types

1.Static balancing:In this only force balancing is done.

2.Dynamic balancing:in this force as well as moment balancing is done.

Balancing become compulsory for over hanging rotor because unbalance force produce lots of vibration and lots of sound due to this rotor or the whole system in which rotor is attached can be damage.

Answer:

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

[tex]1\ gallon = 0.13ft^3[/tex]

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

Answer:

4.186 × 10⁷ J

Explanation:

Heat gain by water = Q

Thus,    

[tex]m_{water}\times C_{water}\times \Delta T=Q[/tex]

For water:  

Volume = 1 m³ = 1000 L ( as 1 m³ = 1000 L)

Density of water= 1 kg/L

So, mass of the water:  

[tex]Mass\ of\ water=Density \times {Volume\ of\ water}[/tex]  

[tex]Mass\ of\ water=1 kg/L \times {1000\ L}[/tex]  

Mass of water  = 1000 kg

Specific heat of water = 4.186 kJ/kg K

ΔT = 10 K

So,

[tex]1000\times 4.186\times 10=Q[/tex]  

Q = 41860 kJ

Also, 1 kJ = 1000 J

So, Q = 4.186 × 10⁷ J

Answer:

36 kg

Explanation:

For water to boil at 105 C it needs a pressure of 121 kPa (this is the vapor pressure of water at 105 C).

In the lid there will be a difference of pressure from one side to the other, this will be compensated by the weight of the lid.

Δp = pwater - patm

Δp = 121 - 101 = 20 kPa

The pressure caused by the weigh of the lid is:

Δp = w / A

Δp = m * g / A

Rearranging

m = Δp * A / g

m = Δp * π/4 * d^2 / g

m = 20000 * π/4 * 0.15^2 / 9.81 = 36 kg

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

Answer:43.70 MPa

Explanation:

Given

mass of engine [tex] 700 lb \approx 317.515 kg[/tex]

diameter of cable [tex]0.375 in.\approx 9.525 mm[/tex]

[tex]A=\frac{\pi d^2}{4}=71.26 mm^2[/tex]

we know stress([tex]\sigma [/tex])[tex]=\frac{load\ applied}{area\ of\ cross-section}[/tex]

[tex]\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa[/tex]

Answer:

Cartesian coordinate system is used to specify any point on a plane.

In two dimensional plane,the two types of axes or coordinates are [tex]x[/tex] and [tex]y[/tex] axis.

Explanation:

Cartesian coordinate system specifies each point and every point on axes.

A Cartesian Coordinate system in two dimension also named as rectangular coordinate system.

The two types of axes or coordinates are [tex]x[/tex] and [tex]y[/tex] axis.

The horizontal axis is [tex]x[/tex]-axis and vertical axis is named as [tex]y[/tex]-axis.

The coordinate system specifies each point as a set of numerical coordinates in a plane which are signed from negative to positive. that is from  ([tex]-\infty[/tex],[tex]\infty[/tex]).

In three dimensional plane, [tex]x[/tex], [tex]y[/tex] and [tex]z[/tex] coordinates are used and in two dimensional plane, [tex]x[/tex] and [tex]y[/tex]coordinates are used to address any point in the interval ([tex]-\infty[/tex],[tex]\infty[/tex]).

For defining both the coordinates, the two perpendicular directed lines [tex]x[/tex]- axis and [tex]x[/tex]-axis are drawn.

Now, for example [tex](3,4)[/tex] is a point in which indicates that the value of [tex]x[/tex] is [tex]3[/tex] and [tex]y[/tex] is [tex]4[/tex].

It is drawn by moving [tex]3[/tex] units right from the origin to positive [tex]x[/tex] axis and [tex]4[/tex] units upwards from the origin [tex](0,0)[/tex] to positive [tex]y[/tex]-axis.

Answer:

a) 152000 slugs

b) 2220000 kg or 2220 metric tons

Explanation:

A body with a weight of 4.9*10^6 lbf has a mass of

4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm

This mass value can then be converted to other mass values.

1 slug is 32.17 lbm

Therefore:

4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs

1 lb is 0.453 kg

Therefore:

4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg

Answer:

Q=1312.5 W

Explanation:

Given that

Cooling load or power input = 750 W

COP=1.75

We know that COP can be given as

COP is the ratio of cooling effect to the input power .

Lets take cooling effect is Q.So now by using COP formula

COP=Q/750

1.75=Q/750

Q=1312.5 W

So the cooling effect produce by air conditioning will be 1312.5 W.

The net effect on laboratory,the laboratory temperature will reduce.