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2 years ago

What is the IMA of this pulley belt system if the diameter of the input

Engineering

1 answer:

Stella [2.4K]2 years ago

8 0

Answer:

2.8

Explanation:

The ideal mechanical advantage of the pulley IMA = D'/D where D' = diameter of output pulley = 7 inches and D = diameter of input pulley = 2.5 inches

So, IMA = D'/D

= 7/2.5

= 2.8

So, the ideal mechanical advantage of the pulley IMA = 2.8

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Draw the ipo chart for a program that reads a number from the user and display the square of that number ???Anyone please

kompoz [17]

Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

Input

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

Processing

Assign variable to the input number (x)

Calculate the square (x = 5 * 5)

Display the result (25) ----> This will be passed to the output module

Output

Display 25

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2 years ago

A freshwater jet boat takes in water through side vents and ejects it through a nozzle of diameter D = 75 mm; the jet speed is V

Radda [10]

Answer:

a) 0.0663 m³/s

b) 3.312 N/(m/s)²

c) 16.665 m/s

d) 0.1105 m³/s

Explanation:

See attached pictures.

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3 years ago

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3 years ago

Read 2 more answers

A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.

AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass $m_{in}+m_{out}=\bigtriangleup m_{system}$

so, $m_e=m_1-m_2$

$Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1$

#Also, given the properties of water as;

$(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg$

#We assume constant properties for the steam at average temperatures: $h_e=\approx(h_1+h_2)/2$

#Replace known values in the equation above; $h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg$

#Using the mass and energy balance relations;

$m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg$

#We have $Q_i_n+m_eh_e=m_2u_2-m_1u_1$ : we replace the known values in the equation as;

$Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ$

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0

3 years ago

Describe three differences between liquids and gases in fluid power systems. Help !!!

scoundrel [369]

Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

3 0

3 years ago