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Tatiana [17]
3 years ago
6

(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg

buddy Bubba has been riding at the front of your 60 kg, 4 m long boat. You come to a stop in the water and then switch places. A) What is the center of mass before you switch places
Physics
1 answer:
valina [46]3 years ago
4 0

Answer:

Explanation:

From the given information:

Let the first weight be m_ 1 = 80 kg

The weight of the buddy be m_2 = 120 kg

The weight of  Bubba be m_3 = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2

The length of the boat be x_2 = 4 m

∴

We can find the center of mass of the system by using the formula:

X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}

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A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
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Answer:

Explanation:

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This force will act in upward direction

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Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

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F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

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3 years ago
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