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Amiraneli [1.4K]
3 years ago
12

when water in a brook or system of pipes flows from a wide region to a narrow region, the speed of water in the narrow region is

Physics
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

more

hope this helps

plz mark brainliest

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A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin
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Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

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dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

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So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

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