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2 years ago

High air pressure systems are usually associated with ________.

Physics

1 answer:

serious [3.7K]2 years ago

7 0

High air pressure systems are usually associated with <u>descending air</u>.

<h3>What does high-pressure air mean?</h3>

Downward motion through the troposphere, the layer of the atmosphere where weather happens, is what creates high-pressure zones. On the other hand, high-pressure systems have higher air pressure than their surroundings. In other words, they are always forcing air away from them and into locations with lower pressure. They are frequently connected to the presence of clear blue skies.

In a high-pressure system, the core experiences a higher pressure than the surrounding regions. From areas of high pressure, winds blow. The winds of a high-pressure system revolve counterclockwise south of the equator and clockwise north of the equator, in the opposite direction from a low-pressure system.

Learn more about high pressure here:

brainly.com/question/12405615

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A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate

PilotLPTM [1.2K]

Answer:

$T(1)=21$

Explanation:

The equation of the position in kinematics is given:

$x(t)=x_{0}+v_{0}t+0.5at^{2}$

x(0) is the initial position, in this it is 0
v(0) is the initial velocity (20 m/s)
a is the acceleration (2 m/s²)

So the equation will be:

$x(t)=20t+0.5*2*t^{2}$

$x(t)=20t+t^{2}$

Now, the Taylor polynomial equation is:

$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

$f(t)=x(t)=20t+t^{2}$

$f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t$

$f''(t)=\frac{dv(t)}{dt}=a(t)=2$

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

$f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}$

$f(0)=x(0)=0$

$f'(0)=v(0)=20$

$f''(0)=a(0)=2$

$T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}$

$T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}$

$T(t)=20t+t^{2}$

Let's put t=1 so find the how far the car moves in the next second:

$T(1)=20*1+1^{2}$

$T(1)=21$

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!

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3 years ago

What is the force on an object with a mass of 25 kg and an acceleration of 5 m/s/s? 1 point

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Answer:125N

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