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GREYUIT [131]
3 years ago
10

1)Which statement best describes energy in an open system?

Physics
2 answers:
zimovet [89]3 years ago
6 0
B an open system flow both 
Aleks04 [339]3 years ago
3 0

the first question is  b) Energy and matter flow into and out of the system. im not sure about the second one

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2 years ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

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3 years ago
A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d
lisabon 2012 [21]

The centripetal acceleration a is 4.32 \times 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m,       Time = 100 s.

Computing the v = 0.4 m / 100 s

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radius of the circular end r = 37 mm = 0.037 m.

            centripetal acceleration a = v^2 / r

                                                        = (4 \times 10^-3)^2 / 0.037

                                                    a = 4.32 \times 10^-4 m/s^2.

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25.5 m/s in 5.75 s. What is the acceleration?
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Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)

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