Ask question

Ask question

All categories

3 years ago

6

A disk‑shaped part is to be cast out of aluminum. The diameter of the disk = 401 mm and its thickness = 25 mm. If the mold const

ant = 1.9 sec/mm2 in Chvorinov's rule, how long will it take the casting to solidify (in the unit of seconds)?

1 answer:

kupik [55]3 years ago

3 0

Answer:

w

Explanation:

You might be interested in

Which allows a user to run applications on a computing device? Group of answer choices Application software CSS Operating system

sveticcg [70]

Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user. The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

Cheers.

4 0

2 years ago

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

2 years ago

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

2 years ago

You are about to perform PMCS on your M1114? What resource should you use for the procedures and instructions for performing PMC

aniked [119]

The resources and instructions that should be used for the procedures of performing PMCS are:

Operator's manuals
Safety cautions and warnings.
Fording kit
Heating and cooling systems.

<h3>What is PMCS?</h3>

PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

M1114

M1151

M1123

Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

Operator's manuals
Safety cautions and warnings.
Fording kit
Heating and cooling systems.

Read more on PMCS here: brainly.com/question/15720250

#SPJ1

4 0

1 year ago

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

$h=\frac{2\sigma cos(\alpha )}{\rho gd}$

where

$\sigma$ is the surface tension of the liquid

$\alpha$ is the contact angle of the liquid

$\rho$ is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

$\alpha =0$

$\rho _{w}=1000kg/m^3$

Applying the values we get

$h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm$

Part b) When the liquid is mercury:

For mercury and glass we have

$\sigma =485.5\times 10^{-3}N/m$

$\alpha =138^o$

$\rho _{w}=13.6\times 10^{3}kg/m^3$

Applying the values we get

$h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm$

The negative sign indicates that there is depression in mercury in the tube.

4 0

2 years ago