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Natali5045456 [20]
1 year ago
14

A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o

f 60℅,to what height will it lift the mass?​
Engineering
1 answer:
navik [9.2K]1 year ago
8 0

The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

useful output energy = 2400 J

Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

h = 2400/800

h = 3 meters

Read more about Machine Efficiency at; brainly.com/question/3617034

#SPJ1

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The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

<h3>​How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

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Read 2 more answers
A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

Here

  • M is the moment which is to be calculated
  • I is the moment of inertia given as

                         I=\frac{bd^3}{12}

Here

  • b is the breath given as 0.75"
  • d is the depth which is given as 8"

                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

So the strain is 8.621 \times 10^{-4} in/in

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

The plastic moment is 600 ksi.

3 0
3 years ago
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