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3 years ago

Starting over, so here's some points! Take em last dayyy ~Jayden

Physics

2 answers:

masya89 [10]3 years ago

3 0

Answer:

Thank you so much!!!! :D

Anit [1.1K]3 years ago

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thank you so much! i really appreciate it

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The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

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pychu [463]

Answer:

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Explanation:

sefamchsgaevfdkahsgevfjahgswevfjahsgefvjashgvf

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3 years ago

If there is 8 g of a substance before a physical change , how much will there be afterwards?

Dimas [21]

A substance undergoing a physical change will still weigh the same even after the change. This is in accordance to the law of conservation of mass which states that mass is neither created nor destroyed. so an 8 g substance remains of the same weight even after undergoing a physical change.

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Answer:

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