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rosijanka [135]
3 years ago
15

A squirrel runs at a steady rate of 0.51 m/s in a circular path around a tree. If the squirrel's centripetal acceleration is 0.4

3 m/s2, what is the radius of the circle?
A.0.60 m
B.1.2 m
C.0.36 m
D.0.84 m
Chemistry
2 answers:
Leokris [45]3 years ago
8 0

Answer:

The radius of the circle is 0.60 meters.

Explanation:

It is given that,

Speed of squirrel, v = 0.51 m/s

Centripetal acceleration, a=0.43\ m/s^2  

Let r be the radius of circle. The expression for centripetal acceleration is given by :

a=\dfrac{v^2}{r}

r=\dfrac{v^2}{a}

r=\dfrac{(0.51\ m/s)^2}{0.43\ m/s^2}

r = 0.60 m

So, the radius of the circle is 0.60 meters. Hence, this is the required solution.

Lelu [443]3 years ago
3 0
I believe It's A. You really should check other sites like before using your points. Just a little pro tip! Another tip: give me brainliest answer pls.
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We can also use the H+ ion concentration to get the hydroxide ion concentration and from that the pOH.

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=========

Just for grins, you might want to know how Kw changes with temperature, and how [H+] and [OH-] are related at some other temperatures. The pH is the pH of a neutral solution at various temperatures. For instance at 10C a neutral solution has a pH of 7.27. That's not a basic pH. 7.27 is the pH of a neutral solution, but at a different temperature. In a neutral solution at 10C [H+] = [OH-] = 5.41x10^-8M.

pH and Kw for a neutral solution at different temperatures

T .........pH ......... Kw

0......... 7.47....... 0.114 x 10-14

10....... 7.27....... 0.293 x 10-14

20....... 7.08....... 0.681 x 10-14

25....... 7.00....... 1.008 x 10-14

30....... 6.92....... 1.471 x 10-14

40....... 6.77....... 2.916 x 10-14

50....... 6.63....... 5.476 x 10-14

100..... 6.14....... 51.3 x 10-14

4 0
3 years ago
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