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Ann [662]
4 years ago
11

A feeding buffer protects ______ path from delays in ______ ____.

Physics
1 answer:
Bad White [126]4 years ago
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>
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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

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

5 0
3 years ago
My car engine can generate 3000 Newtons of force and the car masses 1500 kg. How fast can the sports car accelerate?​
xeze [42]

Answer:

The sports Car can accelerate at 2 m/s².

Explanation:

Force:

Force is the push or pull on an object with mass that causes it to change velocity or to accelerate.

Force represents as a vector, which means it has both magnitude and direction.

Given:

Mass of car = 1500 kg

Force           = 3000 Newtons

To Find:

Acceleration, = ?

Solution:

Formula:

Force = Mass\times Acceleration\\F= ma\\3000 = 1500\times Acceleration\\Acceleration = \frac{3000}{1500} \\Acceleration = 2\ m/s^{2}

Therefore, the sports Car can accelerate at 2 m/s².

8 0
3 years ago
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