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3 years ago

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2 answers:

4 0

A man pulls on a box with 50 N of applied force while another man pushes on the box from the opposite direction with 60 N of applied force. The box will stand still.

TRUE.

barxatty [35]3 years ago

4 0

Answer:

False statement.

Explanation:

It is given that,

Force with which a man pulls a box, F = 50 N (one side)

Force with which pushes on the box, F' = 60 N (in opposite direction)

The forces acting on the box are unbalanced. It means that the force acting in one side is more or less then force in other side. As a result the box will move.

So, the given statement is false.

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2 years ago

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

Illusion [34]

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.

8 0

3 years ago

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