Physics, 05.08.2020 04:01 crtlq

A system of four particles moves along a dimension. The center of mass is at rest, and the particles do not interact with any objects outside of the system. Find the velocity of v4 at t=2.83 seconds given the details for the motion of particles 1,2,3

A system of four particles moves along a dimension. The center of mass is at rest, and the particle

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Answer from: Princess14321

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

$m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg$

The velocity of the center mass of the particles is calculated as;

$M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}$

The velocity of particle 1 at time, t = 2.83 s;

$v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s$

The velocity of particle 2 at time, t = 2.83 s;

$v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s$

The velocity of particle 3 at time, t = 2.83 s;

$v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s$

The velocity of the particle 4 at time, t = 2.83 s;

$v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s$

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

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Answer from: allimaycatp8qgaq

v = - 14.08 m / s

Explanation:

The definition of center of mass is

$x_{cm}$ = 1 /M ∑sun $x_{i} m_{i}$

where M is the total mass of the system and $x_{i}$ and $m_{i}$ are the position and mass of each component.

The velocity of the center of mass can be found by deriving this expression with respect to time

$v_{cm}$ = 1 / M ∑ m_{i} $v_{i}$ vi

let's find the total mass

M = m₁ + m₂ + m₃ + m₄

M = 1.45 + 2.81 +3.89 + 5.03

m = 13.18 kg

let us substitute in the velocity of the center of mass $v_{cm}$ = 0

0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)

v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄

let's substitute the given values

v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03

They ask us for the calculations for a time t = 2.83 s

v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03

v = - 14.08 m / s

Answer from: Quest

b would be the answer

Answer from: Quest

1) when you rub your hands together quickly, the kinetic energy is converted to mechanical energy.2) thermal energy from the sun travels to earth as infrared waves.3) energy is transferred through the matter of space by electromagnetic waves. 4) when a match is burned, chemical potential energy is converted to thermal and radiant energy. 5) seismicwaves are caused by back and forth movement , or vibration, of an object. 6) mechanicalenergy in an engine changes thermal energy into kinetic energy to make a car move.

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A system of four particles moves along a dimension. The center of mass is at rest, and the particles...