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dezoksy [38]
2 years ago
9

vertical gate in an irrigation canal holds back 12.2 m of water. Find the average force on the gate if its width is 3.60 m. Repo

rt your answer with proper units and 3 sig figs.
Engineering
1 answer:
DanielleElmas [232]2 years ago
5 0

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= density\times g\times \frac{h}{2}

By putting values, we get

= 1000\times 9.8\times \frac{12.2}{2}

= 1000\times 9.8\times 6.1

= 59780

hence,

The average force will be:

= Pressure\times Area

= 59780\times 3.6\times 12.2

= 2625537 \ N

Or,

= 2625 \ kN

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assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2
amm1812

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

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3 years ago
What engine does the Mercedes 400e have?
Verdich [7]

Answer:

4.2 m119

Explanation:

it has the 4.2 m119 v8 engine with a top speed of 155 mph and 0 - 60 mph in 6.2 seconds

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3 years ago
What is the best countermeasure against social engineering?
Mkey [24]

Answer:

Hello Monk7294!

Answer:

Employee education

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The most important countermeasure for social engineering is employee education. All the employees should be trained to keep confidential data safe. As a part of security education, organizations have to provide timely orientation about their security policy to new employees. The security policy should address the consequences of the breaches.

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3 0
2 years ago
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blagie [28]

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8 0
3 years ago
A tool chest has 950 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
Mazyrski [523]

Answer:

P > 142.5 N  (→)

the motion sliding

Explanation:

Given

W = 959 N

μs = 0.3

If we apply

∑ Fy = 0 (+↑)

Ay + By = W

If  Ay = By

2*By = W

By = W / 2

By = 950 N / 2

By = 475 N (↑)

Then  we can get F (the force of friction) as follows

F = μs*N = μs*By

F = 0.3*475 N

F = 142.5 N (←)

we can apply

P - F  > 0

P  > 142.5 N (→)

the motion sliding

6 0
3 years ago
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