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Vladimir79 [104]
2 years ago
9

At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at

a constant velocity of 10 m/s in the positive x-direction. At t=3 s, the center of mass of the two objects is at the position x=
Physics
1 answer:
love history [14]2 years ago
4 0

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em>  after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}

x_{cm} = \dfrac{8  \times0 + 16 \times  30}{8 + 16} = 20

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

brainly.com/question/18557256

brainly.com/question/20714030

brainly.com/question/17088562

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Answer: Option B

<u>Explanation:</u>

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                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

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Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

Please hit that thank you button if that helped, I don’t want thank you’s I just want to know that this helped.

Please reply if this doesn’t help, I will try my best to gather more information or a answer.

Here is some good information that could help you out a lot!


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That is some information, I do have more if you need some! Thanks!

Have a great rest of your day/night! :)


Emacathy,
Brainly Team.


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