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2 years ago

Determine the force magnitude p required to lift one end of the 250-kg crate with the lever dolly.

Engineering

1 answer:

Julli [10]2 years ago

8 0

Answer:

224.8125 N

Explanation:

Using the attached diagram, then we can work the answer as follows

Taking moments about 275 mm from point A

$P\times 1500=\frac {mg}{2}\times 275mm$

Where m is the mass, P is the load, g is acceleration due to gravity

$P=\frac {250\times 9.81\times 275mm}{2\times 1500mm}=224.8125 N$

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Water at 400 kPa with a quality of 75% has its pressure raised 200 kPa (to 600 kPa) in a constant volume process. What is the ne

Zolol [24]

Answer:

T=194.3C

X=1

P=542.5kPa

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

First we find the specific volume knowing the temperature and quality of state 1

v(p=400kPa, x=0.75)=0.3471m^3/kg

As a constant volume process is presented, it keeps constant in state 2, so we find the temperature and quality with p = 600kPa and v = 0.3471m ^ 3 / kg

T(P=600kPa, v=0.3471m ^ 3 / kg)=194.3C

x(P=600kPa, v=0.3471m ^ 3 / kg)=1

for the second part we find the pressure, with a quality of 1 and a volume of

v= 0.3471m ^ 3 / kg

p=542.5kPa

4 0

2 years ago

Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in

salantis [7]

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

class Program

{

static void Main(string[] args)

{

int[] Number = new int[10]; // Declare array of 10 elements

//Accept Input

int counter = 0;

while(counter<10)

{

Console.WriteLine("Enter Number " + (counter + 1)+": ");

string var = Console.ReadLine();

Number[counter] = Convert.ToInt32(var);

counter++;

}

//Initialize largest to first element of the array

int Largest = Number[0];

//Determine Largest

for(int i=0;i<10;i++)

{

if(Largest < Number[i])

{

Largest = Number[i];

}

//Print Largest

Console.WriteLine("The largest input is "+ Largest);

Console.ReadLine();

}

8 0

2 years ago

A system consisting of 3 lb of water vapor in a piston–cylinder assembly, initially at 350°F and a volume of 71.7 ft3, is expand

Alla [95]

Answer:

isobaric expansion = 281.09 Btu

isothermal compression= 72 Btu

Explanation:

The first law of thermodynamics is:

$Q_{AB}=W_{AB}+deltaU_{AB}$

where:

Q=heat transferred

W= work

U=internal energy

$W_{AB}=P*(V_{B}-V_{A})$

$U_{AB}=n*C_{v}(T_{2}-T_{1})$

P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.

In a isobaric process heat transferred is:

$Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})$

For an isothermal process (T2-T1 = 0) so

$Q=P*(V_{B}-V_{A})= W_{AB}$

From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.

For the first process

$Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})$

we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)

$specific-volume=\frac{71.7 ft^{3}}{3Lb}=23.9\frac{ft^{3}}{Lb}$

we use this information in the appropriate unit to find the pressure in thermodynamic tables.

T1= 176°C

v1= 1.49 m^3/kg

P=1.37 bar

in the second state we have

P=1.37 bar =137000Pa

$v_{2}=\frac{85.38ft^{3}}{3Lb}= 28.46\frac{ft^{3}}{Lb}$

with thee properties we check in the thermodynamic tables

T2= 255°C

$n=mass/Mw = 3Lb*\frac {1kg}{2.2Lb}*\frac{1000gr}{1kg}*\frac{1mol}{18gr}=75.75 mol$

we usually find Cp on tables for water but from the Mayer relation we have:

$C_{v}=C_{p}+R$

Cp for water vapor is: 33.12 J/mol*K

R=8.314 J/mol*K

Cv= 41.434 J/mol*K

replacing in the equation for Q

$Q=137000 Pa*(2.41m^{3}-2.030m^{3})+75.75mol*41.434\frac{ J}{mol*K}*(528.15-449.81 K)=296569J$

296569J =281.09 Btu

5 0

2 years ago

The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr

Rashid [163]

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below; if we take an integral look at the component of force in the direction of the displacement; we have

$F_x=$ Fcos 40°

$F_x=$ 100 (cos 40°)

$F_x=$ 76.60 N

Workdone by the friction force can now be determined as:

W = $F_x$ × displacement

W = 76.60 × 2

W = 153.2 J

∴ the work done by the friction force = 153.2 J

7 0

2 years ago

A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz

nikitadnepr [17]

Answer:

25 - $\sqrt[4]{26.66*10^{-8} }$ mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2 $\pi$ f = 2 * $\pi$ * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter using the relation

$\frac{J}{c_{2} } = \frac{T}{t_{max} }$ = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

$\frac{J}{c_{2} }$ = $\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )$ = $\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ]$

therefore; 0.025^4 - $c^{4} _{1}$ = 0.050 / $\pi$ (7.8 *10^-6)

$c^{4} _{1}$ = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

$c_{1} = \sqrt[4]{26.66 * 10^{-8} }$ =

THE TUBE THICKNESS

$c_{2} - c_{1}$ = 25 - $\sqrt[4]{26.66*10^{-8} }$ mm

4 0

2 years ago