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2 years ago

What are the 2 main sources of data

Engineering

2 answers:

docker41 [41]2 years ago

8 0

Answer:

I hope this helps

Explanation:

.... Internal source

.... External source

hjlf2 years ago

3 0

Following are the two sources of data:

Internal Source. When data are collected from reports and records of the organisation itself, it is known as the internal source. ...

External Source. When data are collected from outside the organisation, it is known as the external source.

IamSugarBee

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A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

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2 years ago

How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?

nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

= 20.4kg · 9.81 $\frac{N}{kg}$ · 1.50m

= 35.316 $\frac{N}{m}$

= 35.316 W

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Answer:

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Explanation:

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