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Stels [109]
3 years ago
8

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

following code, and visualize the results as a scatter plot: %%sql SELECT cost, price FROM product What should you do before running a cell with the %%sql magic? a. Create a new DataFrame named product from productDF.select("cost", "price") b. Persist the productDF DataFrame using productDF.createOrReplaceTempView("product") c. Filter the productDF dataframe using productDF.filter("cost == price") d. Rename the columns in the productDF DataFrame using productDF.withColumnRenamed("cost", "price")
Engineering
1 answer:
defon3 years ago
7 0
You need to explain it more simple as everyone is clueless
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A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying
Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius,r_{2} = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, r_{1} = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

 τmax  = \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

3 0
4 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
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