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4 years ago

A feeding buffer protects path from delays in .

1 answer:

7 0

<span> A feeding buffer protects critical path from delays in non-Critical path

Basically, a feeding buffer acts as a protection for the critical chains from the possible violations in the feeding chains.
By doing this, you would add a protection to the baseline of the deadline of your current projects</span>

You might be interested in

Which of these should you always do at the end of a calculation

yulyashka [42]

Answer:

Reverse check the answer

Explanation:

I believe it is very important that once someone is done with any calculation, the person ought to go over the calculations again. And even, recheck the answer in inverted form.

This is so because while doing the calculations, we can possibly make errors that we won't notice until after submission. Knowing 2 * 3 = 6, but writing 2 * 3 = 5 in the course of calculations can happen to anybody. So therefore, cross checking and reverse checking is needed

8 0

3 years ago

Pls answer my quetion i will give brainlist answer

allochka39001 [22]

Answer:

here's an explanation but not the answer

please kindly send No or Yes understand.

Explanation:

A 10-N force is applied to push a block across a friction free surface for a displacement of 5.0 m to the right.

See Answer

A 10-N frictional force slows a moving block to a stop after a displacement of 5.0 m to the right.

7 0

3 years ago

A thin stream of water flows smoothly from a faucet and falls straight down. At one point the water is flowing at a speed of v1

MrRa [10]

Answer:

The vertical distance between these two points is 12.28 cm.

Explanation:

Given that,

Speed of water = 1.23 m/s

Diameter at lower point d'= 0.787 d

We need to calculate the speed of water

Using continuity equation

$A_{1}v_{1}=A_{2}v_{2}$

$\dfrac{\pi\times d^2}{4}v_{1}=\dfrac{\pi\times d'^2}{4}v_{2}$

$v_{2}=v_{1}\times(\dfrac{d}{d'})^2$

Put the value into the formula

$v_{2}=1.23\times(\dfrac{d}{0.787d})^2$

$v_{2}=1.98\ m/s$

We need to calculate the vertical distance h between these two points

Using Bernoulli theorem

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}$

Here, P₁ = P₂ = atmosphere pressure is same because both end is open

$\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)=\rho\times g(h_{1}-h_{2})$

$(v_{2}^2-v_{1}^2)=2g(h_{1}-h_{2})$

$2g\times\Delta h=(v_{2}^2-v_{1}^2)$

$\Delta h=\dfrac{1.98^2-1.23^2}{2\times9.8}$

$\Delta h=12.28\ cm$

Hence, The vertical distance between these two points is 12.28 cm.

6 0

4 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C