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3 years ago

Help when u can! I feel like it's C, but I just want to make sure before I submit the test.

Physics

1 answer:

Marat540 [252]3 years ago

4 0

(D)

Explanation:

This kind of bond where the electrons are transferred is called ionic bond. This bond occurs between a metallic and non-metallic elements. The other kind of bond (covalent) involves the sharing of electrons and this happens between non-metallic elements.

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Ball A 1.55kg moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B m

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1.15 m/s to the left (3 sig. fig.).

<h3>Explanation</h3>

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

$p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}$

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ , where

$m$ stands for mass and
$v$ stands for velocity, which can take negative values.

Let the velocity of objects moving to the right be positive.

$m_\text{A} = 1.55\;\text{kg}$ ,
$m_\text{B} = 0.752\;\text{kg}$ .

Before the two balls collide:

$v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1}$ ,
$v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}$ .

After the two balls collide:

$v_\text{A}$ needs to be found,
$v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}$ .

Again,

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ ,

$1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03)$ .

$v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}$ .

$v_{\text{A,final}}$ is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.

Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.

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