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iren2701 [21]
4 years ago
12

"The team wants to delete from Dataset A all activities related to users in Dataset B. How can an Einstein Consultant help them

achieve this_______?
Physics
1 answer:
lesantik [10]4 years ago
5 0

Adittional Information:

The Universal Containers company used Einstein Analytics to create two datasets:

Dataset A: contains a list of activities with an "activitylD" dimension and a "userlD" dimension

Dataset B: contains a list of users with a "userlD" dimension

Answer Choices:

A. Use the dataflow transformation "delete" and set "userlD" as the deletion ID.

B. Use a combination of dataflow transformations: "augment" and "filter."

C. Use an external ETL tool to extract both datasets and delete records.

D. Use the recipe operation "delete" and set "userlD" as the deletion ID.

B. Use a combination of dataflow transformations: "augment" and "filter."

Answer:

B. Use a combination of dataflow transformations: "augment" and "filter."

You might be interested in
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc
lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

E =  \frac{V}{L}

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

J = \frac{E}{p}

Substitute (V/L)  for E in the above equation of current density.

J = \frac{V}{pL} ------(1)

Substitute iR for V in equation (1)

J = \frac{iR}{pL} ------(2)

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

p = \frac{RA}{L}

A = \frac{pL}{R}

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A  and λ is use for (m/L) in the eqn above

\lambda = d\frac{p}{\frac{R}{L} } ------(3)

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³  for d and 1.69 × 10⁸ Ω .m

\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1

c) Using the equation (2) current density for aluminum cable is,

J = \frac{iR}{pL}

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

\lambda = d\frac{p}{\frac{R}{L} }

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0
3 years ago
Read 2 more answers
Water flows through a horizontal nozzle in steady flow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2
Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

        Q = A v

        v = Q / A

The area of ​​a circle is

       A = π r² = π d² / 4

Let's look at the speeds at each point

       v₁ = Q / A₁ = Q 4 /π d₁²

       v₁ = 10 4 /π 0.5²

       v₁ = 50.93 m / s

       v₂ = Q / A₂

       v₂ = 10 4 /π 0.25²

       v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

       P₁ = P2 + ½ rho (v₂² - v₁²)

      P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

      P₁ = 1.013 10⁵ + 2.205 10⁷

      P₁ = 2.215 10⁷ Pa

la definicion de presion es

      P₁ = F₁/A₁

     F₁ = P₁ A₁

     F₁ = 2.215 10⁷ pi d₁²/4

     F₁ = 2.215 10⁷ pi 0.5²/4

     F₁ = 4.3 106 N

     

6 0
3 years ago
!!!!HURRY 30 POINTS!!!!
Nataly_w [17]
The anwser is C flooding
3 0
3 years ago
A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specif
forsale [732]

Answer:

500cal

Explanation:

Given parameters:

Mass of water  = 50g

Initial temperature  = 22°C

Final temperature  = 32°C

Specific heat of water  = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories  = ?

Solution:

To solve this problem, we use the expression below:

       H  = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

       H  = 50 x 1 x (32  - 22)  = 500cal

5 0
3 years ago
A 10n falling object encounters 4n of air resistance. what is the net force on the object?
Margaret [11]
It would be 6n down.
5 0
3 years ago
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