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Nastasia [14]
3 years ago
6

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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because lighting heats up the surrounding air slowly, it realationship to thunder is currently unknown true or false​
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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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