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rosijanka [135]
4 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]4 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]4 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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Gas is heated from 480. K to 750. K and the pressure is kept constant, what final volume would result if the original volume was
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4 0
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using information in titles of appendices b and c, calculate the minimum grams of propane, C3H8 (g), that must be combusted to p
Lubov Fominskaja [6]

<u>Given:</u>

Mass of ice = mass of water = 5.50 kg = 5500 g

Temperature of ice = -20 C

Temperature of water = 75 C

<u>To determine:</u>

Mass of propane required

<u>Explanation:</u>

Heat required to change from ice to water under the specified conditions is:-

q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)

  = m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)

  = 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ

The enthalpy change for the combustion of propane is -2220 kJ/mol

Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane

Molar mass of propane = 44 g/mol

Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g

Ans: 75.15 grams of propane must be combusted.



4 0
4 years ago
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