Who plays blox burg???

Which option identifies the next step in the following scenario?

Whitepunk [10]

Answer: The engineer will create a detailed sketch that labels all of the visual components.

Explanation:

It should be noted that the reverse engineering is required for the replacement and the modification of an existing product.

With regards to the question, the correct answer is option A "The engineer will create a detailed sketch that labels all of the visual components".

4 0

3 years ago

What’s the number of gold atoms in a nanogram? a picogram?

zvonat [6]

Answer :

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

The number of gold atoms in picogram is, $3.057\times 10^{9}$

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-9}$ nanograms of gold contains $\frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12}$ number of gold atoms

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-12}$ picograms of gold contains $\frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9}$ number of gold atoms

The number of gold atoms in picogram is, $3.057\times 10^{9}$

8 0

3 years ago

Given x = 67 and y = 18, use two's complement to calculate:(1)x+y (2) x-y (3)-x+y (4)-x-y. Do the calculation, show the results

joja [24]

Answer:

See attachment

Explanation:

8 0

3 years ago

With a very precise volumetric measuring device, the volume of a liquid sample is determined to be 6.321 L (liters). Three stude

zheka24 [161]

Answer:

See explanation

Explanation:

Solution:-

- Three students measure the volume of a liquid sample which is 6.321 L.

- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:

Students

Trial A B C

1 6.35 6.31 6.38

2 6.32 6.31 6.32

3 6.33 6.32 6.36

4 6.36 6.35 6.36

- We will define the two terms stated in the question " precision " and "accuracy"

- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.

- The mean measurement taken by each student would be as follows:

$E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\$

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

$Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\$

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:

Var ( A ) < Var ( B ) < Var ( C )

most precise Least precise

- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

$P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\$

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:

P ( B ) < P ( A ) < P ( C )

most accurate least accurate

7 0

3 years ago

Determine ten different beam loading values that will be used in lab to end load a cantilever beam using weights. Load values sh

nasty-shy [4]

Answer:

1st value = 1.828 * 10 ^9 gm/m^2 ------- 10th value = 7.312 * 10^9 gm/m^2

Explanation:

initial load ( Wp) = 200 g

W1 ( value by which load values increase ) = 100 g

Ten different beam loading values :

Wp + w1 = 300g ----- p1

Wp + 2W1 = 400g ---- p2

Wp + 3W1 = 500g ----- p3 ----------------- Wp + 10W1 = 1200g ---- p10

x = 10.25" = 0.26 m

b = 1.0" = 0.0254 m

t = 0.125" = 3.175 * 10^-3 m

using the following value to determine the load values at different beam loading values

attached below is the remaining part fo the solution

5 0

3 years ago