a box half full of apples weighs 40kg and the same box one fifth full weighs 22kg, what does the box weigh when empty?

2 answers:

6 0

1/2 = 40kg
1/5 = 22kg
1/2 = 5/10
1/5 = 2/10
3/10 is the difference between 18kg.
This means 1/10 of the box filled with apples is 6kg.
<span>40 - (5 x 6) = 10kg. The box weighs 10kg alone. </span>

ruslelena [56]4 years ago

5 0

1/2 = 40kg
1/5 = 22kg
1/2 = 5/10
1/5 = 2/10
3/10 is the difference between 18kg.
This means 1/10 of the box filled with apples is 6kg.
40 - (5 x 6) = 10kg. The box weighs 10kg alone.

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$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

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Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$