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3 years ago

10

Determine the magnitude of force P needed to start towing the 40kg crate.Also determine the location of the resultant normal for

ce acting on the crate

1 answer:

erastova [34]3 years ago

5 0

The distance from Point A=500 mm

Explanation:

M := 40kg c := 200mm

μs := 0.3 d := 3

a := 400mm e := 4

b := 800mm

Initial guesses: $N_{c}$ := 200 N P := 50N

Plz refer to the image below

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Explain why it is not advisable to wear Ornament like ring at work shop

disa [49]

The reason that it is not advisable to wear Ornament like ring at work shop is that rings can be easily get caught and fingers or the hands of the person can be injured, cut, scared etc.

<h3>Why should you not wear a ring at work?</h3>

Jewelry is known to bring about a lot of safety hazards for people working around chemicals and others.

Note that the reason that it is not advisable to wear Ornament like ring at work shop is that rings can be easily get caught and fingers or the hands of the person can be injured, cut, scared etc.

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2 years ago

Steam enters a turbine in a Rankine cycle power plant at 200 psia and 500 °F. a) Calculate the isentropic thermal efficiency if

Aleks04 [339]

Answer:

η=0.19=19% for p=14.7psi

η=0.3=30% for p=1psi

Explanation:

enthalpy before the turbine, state: superheated steam

h1(p=200psi,t=500F)=2951.9KJ/kg

s1=6.8kJ/kgK

Entalpy after the turbine

h2(p=14.7psia, s=6.8)=2469KJ/Kg

Entalpy before the boiler

h3=(p=14.7psia,x=0)=419KJ/Kg

Learn to pronounce

the efficiency for a simple rankine cycle is

η= $\frac{h1-h2}{h1-h3}$

η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)

η=0.19=19%

second part

h2(p=1psia, s=6.8)=2110

h3(p=1psia, x=0)=162.1

η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)

η=0.3=30%

7 0

3 years ago

Select the correct answer.

Brilliant_brown [7]

The answer is either b or e search it up and you will know the answer by definition

4 0

3 years ago

Read 2 more answers

Alternating current lesson 4 exam

zloy xaker [14]

Huh? whats the question?

6 0

2 years ago

A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 s

Nezavi [6.7K]

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) = 2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

6 0

3 years ago