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bearhunter [10]
3 years ago
15

Anything you want to do in Hootsuite can be found in the ________, with the main workspace in the _________?

Engineering
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

Anything you want to do in Hootsuite can be found in the ___ Sidebar_____, with the main workspace in the ___center______

Sidebar; center

Explanation:

The main workspace of the Hootsuite is located in the center.  The sidebar is where the core access to the Hootsuite functionality, like Streams, Inbox, Planner, Analytics, Publisher, and the App Directory, is obtained.  Hootsuite is a media management platform for curating content, scheduling posts, managing team members, and measuring performances.

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Impedance is defined as the total opposition to current in an AC circuit. Question 17 options: True False
ArbitrLikvidat [17]

Answer: True

Explanation:

The total opposition to current flow in an AC circuit is known as Impedance.

8 0
2 years ago
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co
butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

5 0
3 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
Alex Ar [27]

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

4 0
3 years ago
Risks are Not Perceived Differently from What is Happening<br> False<br> True
Gnoma [55]

Answer:

False......................

8 0
3 years ago
How could increasing the budget for testing have prevented the problem experienced by the mars orbiter?
vlada-n [284]

Answer:

this might help

Explanation:

https://science.ksc.nasa.gov/mars/msp98/misc/MCO_MIB_Report.pdf

5 0
3 years ago
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