What is the range of a 32-bit signed integer?

Answer:

(A) Dimension of force  [tex]MLT^{-2}[/tex]

(b) Dimension of energy [tex]ML^2T^{-2}[/tex]

(c) Dimension of power = [tex]ML^2T^{-3}[/tex]

Explanation:

We have to dimension

(a) Force

We know that force F = ma

Dimension of mass is M, and dimension of a is [tex]LT^{-2}[/tex]

So dimension of force [tex]MLT^{-2}[/tex]

(B) Energy

We know that energy = force × displacement

In previous part we can see dimension of force is   [tex]MLT^{-2}[/tex]

and dimension of displacement = L

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

(c) Power

We know that power [tex]p=\frac{energy}{time }[/tex]

In previous part we can see dimension of energy is [tex]ML^2T^{-2}[/tex]

And dimension of time is T

So  dimension of power = [tex]ML^2T^{-3}[/tex]

Explanation:

Step1

Factor of safety is the constant factor which is taken for the safe design of any product. It is the ratio of maximum stress induced in the material or the failure stress from tensile test to the allowable stress.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the type of material, environment condition, strength needed, type of component etc.

The expression for factor of safety is given as follows:

[tex]FOS=\frac{\sigma_{f}}{\sigma_{a}}[/tex]

Here, [tex]\sigma_{f}[/tex] is fracture stress and [tex]\sigma_{a}[/tex] is allowable stress.

Answer:

The flow is turbulent at both the parts of the tube.

Explanation:

Given:  

Water is flowing in circular tube.  

Inlet diameter is [tex]d_{1}= 2[/tex]m.  

Outlet diameter is [tex]d_{2}= 3[/tex]m.  

Inlet velocity is [tex]V_{1}= 3[/tex] m/s.  

Kinematic viscosity is [tex]\nu =1.24\times  10^{-6}[/tex] m²/s.  

Concept:

Apply continuity equation to find the velocity at outlet.  

Apply Reynolds number equation for flow condition.  

Step1  

Apply continuity equation for outlet velocity as follows:  

[tex]A_{1}V_{1}=A_{2}V_{2}[/tex]  

[tex]\frac{\pi}{4}d^{2}_{1}V_{1}=\frac{\pi}{4}d^{2}_{2}V_{2}[/tex]  

Substitute the values in the above equation as follows:  

[tex]\frac{\pi}{4}2^{2}\times 3=\frac{\pi}{4}3^{2}V_{2}[/tex]  

[tex]2^{2}\times 3=3^{2}V_{2}[/tex]  

[tex]V_{2}=\frac{4}{3}[/tex] m/s.  

Step2

Apply Reynolds number formula for the flow condition at inlet as follows:  

[tex]Re=\frac{v_{1}d_{1}}{\nu }[/tex]  

[tex]Re=\frac{2\times 3}{1.24\times  10^{-6}}[/tex]  

Re=4838709.677  

Apply Reynolds number formula for the flow condition at outlet as follows:  

[tex]Re=\frac{v_{2}d_{2}}{\nu }[/tex]  

[tex]Re=\frac{\frac{4}{3}\times 3}{1.24\times 10^{-6}}[/tex]  

Re=3225806.452  

Thus, the Reynolds number is greater than 2000. Hence the flow is turbulent.  

Hence, the flow is turbulent at both the parts of the tube.  

Answer:

P=7.027psi

V=1.9788ft ^3

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.

For the first part of this problem, we must find the water saturation pressure at a temperature of 177F

Psat @(177F)=7.027psi

For the second part of this problem, we calculate the specific volume of the water knowing the temperature and that the state is saturated liquid, then multiply by the mass to know the volume

v@177F=0.01649ft^3/lbm

V=mv

V=(0.01649ft^3/lbm)(120lbm)

V=1.9788ft ^3

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

To calculate the total lb mol air/h needed to heat the hydrocarbon oil, we can use the principle of heat transfer. First, calculate the heat transfer between the hot air and the oil using the formula Q = mcΔT. Finally, divide the heat transfer by the heat capacity of air to find the total lb mol air/h needed.

To calculate the total lb mol air/h needed to heat the hydrocarbon oil, we can use the principle of heat transfer. First, we need to calculate the heat transfer between the hot air and the oil using the formula Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the temperature difference. We know that the heat exchanger transfers 2200 lbm/h of oil from 100°F to 150°F, so the total heat transfer is Q = 2200 lbm/h * (150°F - 100°F) * 0.45 btu/lbm • °F. Next, we can calculate the lb mol of air needed by dividing the heat transfer by the heat capacity of air, which is 0.24 btu/lbm • °F. Therefore, the total lb mol air/h needed is Q / (0.24 btu/lbm • °F).

Answer:

The main component of fixed wing aircraft

1.Wings

2.Fuselages

3.Landing gear

4.Stabilizers

5.Flight control surface

Cantilever wing:

  A cantilever wing is directly attached to the fuselages and do not have any external support.

Semi cantilever wing:

A cantilever wing does not directly attached to the fuselages and  have any external support.The external support may be one two.

Answer:

1) A=282.6 mm

2)[tex]a_{max}=60.35\ m/s^2[/tex]

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

[tex]V=\omega \sqrt{A^2-x^2}[/tex]

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]    ------3

[tex]2.5=\omega \sqrt{A^2-0.225^2}[/tex]   --------4

Now by dividing equation 3 by 4

[tex]\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}[/tex]

[tex]1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}[/tex]

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]

[tex]3.5=\omega \sqrt{0.2826^2-0.15^2}[/tex]

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

[tex]a_{max}=\omega ^2A[/tex]

[tex]a_{max}=14.61 ^2\times 0.2826\ m/s^2[/tex]

[tex]a_{max}=60.35\ m/s^2[/tex]

Time period T

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]T=\dfrac{2\pi}{14.609}[/tex]

T=0.42 sec

The question involves calculating the duration fuel will last in a power station with 12% efficiency given its energy content and the daily energy demand. A misunderstanding in the conversion of units and efficiency calculation was identified in the attempt, highlighting the need for a corrected mathematical approach involving unit conversions and understanding power station efficiency.

A certain amount of fuel contains 15×102 Btu of energy and is converted into electric energy in a power station having a 12 percent overall efficiency. The average daily demand on the station is 5MW. To find out in how many days the fuel will be totally consumed, first, it's essential to calculate the total energy converted into electrical energy considering the efficiency of the power station. Given the power station efficiency, only 12% of the fuel energy is converted into electrical energy. The total energy available for conversion is 15×102 Btu.

First, convert 15×102 Btu into megajoules (MJ) as 1 Btu is approximately equal to 0.00105506 MJ. Therefore, the energy in MJ is 15×102 × 0.00105506 = 1.58259 MJ. Given that 12% of this energy is usable for electricity, the usable energy is 1.58259 MJ × 12% = 0.189911 MJ. Since 1 MW is 1 MJ/sec, and given the station's demand is 5MW over 24 hours, the total consumption is 5MW × 24h = 120 MW/day.

Then, to find out how many days the fuel would last, divide the total usable electrical energy by the daily consumption. However, an error in the calculation process prevents us from completing the result as intended initially. This calculation seems to involve a misunderstanding regarding the conversion of units and the calculation of efficiency. Therefore, a corrected approach should involve accurately converting the initial Btu to MJ or kWh, taking into account the efficiency correctly, and then calculating the duration of fuel consumption based on the daily energy output and demand.

Answer:

Time taken by the [tex]1\mu m[/tex] diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = [tex]1\times 10^{- 6} m[/tex]

Diameter of the smaller droplet, R' = [tex]0.5\times 10^{- 6} m[/tex]

Now,

Volume of the droplet, V = [tex]\frac{4}{3}\pi R^{3}[/tex]

Volume of the smaller droplet, V' = [tex]\frac{4}{3}\pi R'^{3}[/tex]

Volume of the droplet ∝ Time taken for complete evaporation

Thus

[tex]\frac{V}{V'} = \frac{t}{t'}[/tex]

where

t' = taken taken by smaller droplet

[tex]\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}[/tex]

[tex]\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}[/tex]

t' = [tex]60\times 10^{- 9} s = 60 ns[/tex]

Answer:

12024000 lb*ft

Explanation:

The total work will be the sum of the energy consumed by friction plus the kinetic energy the car attained.

L = Ek + Lf

Lf = Ff * d

Ek = 1/2 * m * v^2

Therefore:

L = Ff * d + 1/2 * m * v^2

L = 80 * 300 + 1/2 * 15000 * 40^2 = 12024000 lb*ft

THe total work done on the car is of 12024000 lb*ft

Answer:

P=41.84psi

Explanation:

Cavitation is a phenomenon that arises when the pressure of a liquid fluid reaches the vapor pressure, this means that if at any point in a system of pipes and pumps the water pressure is equal to or less than the vapor pressure it begins to evaporate, producing explosive bubbles that can cause damage and noise in the system.

Therefore, for this problem, all we have to do is find the water vapor pressure at a temperature of 270F, using  thermodynamics tables

Pv@270F=41.84psi

Answer:

[tex]p = 15260.643 \ lbf/ft^2[/tex]

Explanation:

person weight is 175 lbm

weight of stake board 4lbm

size of stakeboard = 2ft by 0.8 ft

area of stakeboard is [tex]2*0.8 ft^2 = 1.6 ft^2[/tex]

gauge pressure is given as

[tex]p =\frac{ w_p+w_s}{A} g[/tex]

where is g is acceleration due to gravity =  32.17 ft/sec^2

puttng all value to get pressure value

   [tex]= \frac{175 +4}{1.6}* 32.17[/tex]

[tex]p = 15260.643 \ lbf/ft^2[/tex]

Answer:

Determine A) The Volume Of The Tank (ft^3) Later A Pump Is Used To Extract ... A rigid, sealed cylinder initially contains 100 lbm of water at 70 degrees F and atmospheric pressure. ... Later a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibrium ...

Answer:

FALSE.

Explanation:

the correct answer is FALSE.

Projection is the process of representing the 3 D object on the flat surface.

there are four ways of representing the projection

1) First angle projection  

2) second angle projection

3) third angle projection

4) fourth angle projection.

Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.

In USA people uses the third angle of projection.

Answer:

The pressure in pounds per square inch (psi) equals 14.50 psi.

Explanation:

Since psi stands for 'pounds per square inch' and Pascals stands for 'Newtons per square meters'

We can write the given pressure as

[tex]Pressure=100kPa\\\\Pressure=100\times 10^{3}N/m^{2}[/tex]

Now since 1 Newton of force equals 0.22481 Pound force also

since 1 meter equals 39.37 inches

Using the above values we get

[tex]Pressure=100\times 10^{3}\times \frac{0.22481lb}{(39.37)^{2}inch^{2}}[/tex]

thus the pressure becomes

[tex]Pressure=100\times 10^{3}\times \frac{0.22481}{(39.31)^{2}}pounds/inch^{2}\\\\Pressure=14.50psi[/tex]

Answer:

[tex]\Delta T = 28.57°C[/tex]

Explanation:

given data:

temperature at outlet of turbine = 260°C

Flow rate = 2 kg/min = 0.033 kg/s

output power  = 1 kW

specific heat =  1.05 kJ/kg.K

We know that power generated by turbine is equal to change in enthalpy

[tex]W = h_2 - h_1[/tex]

   [tex]= mCp(\Delta T)[/tex]

[tex]1*10^3 = 0.0333*1.05*10^3 * \Delta T[/tex]

[tex]\Delta T = \frac{10^3}{0.033*1.05*10^3}[/tex]

[tex]\Delta T = 28.57°C[/tex]

Answer:

a)True

Explanation:

In cutting action,three type of zone are presents

1.Primary zone :

The most of part of energy is converted in to heat.

2.Secondary zone:

Heat generation due to rubbing between tool and chip.This also called deformation zone.

3.Tertiary zone:

Heat generated due to flank of tool and already machined surface.

So the highest temperature is located close to the shear zone.

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

[tex] {v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2} [/tex]

Now, we substitute v, u, and a in the formula

[tex]v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds[/tex]

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Answer:

The modulus of toughness is greater for ductile material.

Explanation:

Modulus of toughness is defined as the amount of strain energy that a material stores per unit volume. It is equal to the area under stress-strain curve of the material up to the point of fracture.

As we know that  the area under the stress-strain curve of a ductile material is much more than the area under the stress strain curve of a brittle material as brittle materials fail at a lesser load hence we conclude that the modulus of toughness is greater for ductile material than a brittle material.

In our experience we can also relate that for same volume a substance such as glass sheet (brittle) fails at lower load as compared to a sheet of steel (ductile) with identical dimensions.

Answer:

Diameter of riser =6.02 mm

Explanation:

Given that

Dimensions of rectangular plate is 200mm x 100mm x 20mm.

Volume of rectangle V= 200 x 100 x 20 [tex]mm^3[/tex]

Surface area of rectangle A

A=2(200 x 100+100 x 20 +20 x 200)[tex]mm^2[/tex]

So V/A=7.69

We know that

Solidification times given as

[tex]t=K\left(\dfrac{V}{A}\right)^2[/tex]   -----1

Lets take diameter of riser is d

Given that riser is in spherical shape so V/A=d/6

And

Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.

Lets take [tex]\dfrac{V}{A}=M[/tex]

[tex]\dfrac{M_{rac}}{M_{riser}}=\dfrac{7.69}{\dfrac{d}{6}}[/tex]

Now from equation 1

[tex]\dfrac{3.5}{4.2}=\left(\dfrac{7.69}{\dfrac{d}{6}}\right)^2[/tex]

So by solving this d=6.02 mm

So the diameter of riser is 6.02 mm.

Answer:

a) 100%

b) 300%

c) 301 %

Explanation:

The first wafer has a diameter of 150 mm.

The second wafer has a diameter of 300 mm.

The second wafer has an increase in diameter respect of the first of:

((300 / 150)  - 1) * 100 = 100%

The first wafers has a processable area of:

A1 = π/4 * D1^2

The scond wafer has a processable area of:

A2 = π/4 * D2^2

The seconf wafer has a increase in area respect of the first of:

(A2/A1 * - 1) * 100

((π/4 * D2^2) / (π/4 * D1^2) - 1) * 100

((D2^2) / (D1^2) - 1) * 100

((300^2) / (150^2) - 1) * 100 = 300%

The area of a chip is

Ac = Lc^2

So the chips that can be made from the first wafer are:

C1 = A1 / Ac

C1 = (π/4 * D1^2) / Lc^2

C1 = (π/4 * 150^2) / 10^2 = 176.7

Rounded down to 176

The chips that can be made from the second wafer are:

C2 = A2 / Ac

C2 = (π/4 * D2^2) / Lc^2

C2 = (π/4 * 300^2) / 10^2 = 706.8

Rounded down to 706

The second wafer has an increase of chips that can be made from it respect of the first wafer of:

(C2 / C1 - 1) * 100

(706 / 176 - 1) *100 = 301%

Answer:

The pressure will be measured to a precision of 0.073 psi.

Explanation:

Since the relation between the measurement of pressure and height is given by

[tex]dP=\rho gh[/tex]

For water we have

[tex]\rho _{water}=62.4lb/ft^3[/tex]

[tex]g=32.17ft/s^2[/tex]

Applying the given values we get

[tex]dP=62.4\times 32.17\times \frac{1}{16\times 12}=5.433lb/ft^2\\\\=\frac{10.455}{144}lb/in^2=0.073psi[/tex]

Answer:

parallel ; Viscosity

Explanation:

Laminar is flow is the flow of fluid layer in which motion of the liquid particle is very slow and there is no intermixing of the layer of fluid takes place.

There no perpendicular movement of particles takes place, no eddies are formed or swirl of fluid.

so, the first option will be Parallel. Fluid particles flow parallel to each other in laminar flow.

This path of flow is by the action of Viscosity of the fluid.

Answer:

13282.3 years

Explanation:

The C-14 decays exponentially:

[tex]\frac{dN}{dt} =λ*N[/tex]

The solution for this equation is  

[tex]N= N_{o}*e^{- λt}[/tex]

Where:  

No = atom number of C-14 in t=0

N = atom number of C-14 now  

I= radioactive decay constant  

clearing t this equation we get:  

[tex]t=-\frac{1}{ λ}*ln\frac{N}{N_{o}}[/tex]

The term 1/I is called half-life and the value for C-14 is 8252 years.  

N for this exercise is 0.2No  

[tex] t= -8033 * ln \frac{0.2N_{o} }{N_{o}}[/tex]

t = 13282.3 years

Answer:

The large percentage of steel includes less than 0.35% carbon

Explanation:

Carbon is perhaps the most important business alloy of steel. Raising carbon material boosts strength and hardness and enhances toughness. However, carbon often increases brittleness.

The large percentage of steel includes less than 0.35% carbon. Any steel with a carbon content range of 0.35% to 1.86% can be considered as hardened.

Explanation:

Mid-wing configuration places wings exactly at midline of airplane which means at half of height of fuselage. These airplanes are well balanced and also they have a large control surface area.It is the best option aerodynamically as these planes are streamlined much more and also has low interference drag as compared to the high and the low wing configurations.

The mid-wing has almost neutral roll stability that is further good from prespective of the combat as well as the aerobatic aircraft as mid-wing allows for performance of the rapid roll maneuvers with the minimum yaw coupling.

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

Answer:

If the shearing stress is linearly related to shearing strain then the fluid is called as Newtonian fluid.

Explanation:

The other types of fluids are:

1) Non-Newtonian fluids which are further classified as

a) Thixotropic Fluid: Viscosity decreases with shearing stress over time.

b) Rehopactic Fluid: Viscosity increases with shearing stress over time.

c) Dilatant Fluid: Apparent viscosity increases with increase in stress.

d) Pseudoplastic: Apparent viscosity decreases with increase in stress.

Answer:

If the shearing stress is linearly related to shearing strain then the fluid is called as Newtonian fluid.

Explanation:

Explanation:

Step1

Float glass is the process of glass manufacturing on the flat surface of metal like tin. In this method molten glass is allowed to float on the surface of metal.

Step2

Float glass gives the uniform and flat surface of glass product. The thickness of the glass produced is uniform throughout. This process of glass making is very cheap and has negligible distortion. Flat glass process is called the float process because of producing high quality flat surface.

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

finally you use the two previous equations to make a system and find the mass flows

I attached procedure