What is the least count of screw gauge? (a) 0.01 cm (b) 0.001 cm (c) 0.1 cm (d) 1 mm

Which two is right about febuary 14

igor_vitrenko [27]

Answer:A and B

Explanation:

3 0

3 years ago

Read 2 more answers

Write a function named "total_population" that takes a string then a list as parameters where the string represents the name of

adelina 88 [10]

Answer:

Return the total population of all cities in the list.

Explanation:

It is for every element in cityinfo. It works not only for one array but multiple.

I attached the document with the code or function with the name import cvs becuase when I wrote it down and it sent a message written on red about inappropriate words.

Download docx

5 0

3 years ago

Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each

Elina [12.6K]

The initial void ratio is the parameter which is used to show the structural foundations for each specimen of sand so that the method and speed of compression would be measured.

Relative density is the mass per unit volume of each specimen of sand which is measured and it has to do with the relative ratio of the density of the sand.

Unit weight is the the exact weight per cubic foot of the sand which is measured.

Please note that your question is incomplete so I gave you a general overview to help you better understand the concept

Read more here:

brainly.com/question/15220801

5 0

2 years ago

Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

5 0

3 years ago