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4 years ago

If you exert a force of 100.0 N to lift a box a distance of 0.5 m, how much work do you do? 200 J 400 J 50 J 26 J

Physics

1 answer:

jeyben [28]4 years ago

4 0

Work = force x distance
= 100N (force) x 0.5m (distance)
= 50J

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A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

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How are mixtures essential to living and non living things on earth?

sertanlavr [38]

Let's mention 2 examples of a mixture:

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andrew-mc [135]

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Ignoring air resistance, an object in free fall will fall d feet in t seconds, where d and t are related by the algebraic mode

Lostsunrise [7]

Explanation:

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