A plank with a box on it at one end is gradua | vedclass

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(C) $1$. Coefficient of static friction $(\mu_s)$: When the box just starts to slip,the angle of inclination is the angle of repose. Thus,$\mu_s = 	a

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$0.6$ and $0.5$

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(C) $1$. Coefficient of static friction $(\mu_s)$: When the box just starts to slip,the angle of inclination is the angle of repose. Thus,$\mu_s = 	an 	heta = 	an 30^o = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.6$.$2$. Coefficient of kinetic friction $(\mu_k)$: The acceleration $a$ of the box sliding down the plank is given by $a = g \sin 	heta - \mu_k g \cos 	heta$.$3$. Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = 4.0\, m$,$u = 0$,and $t = 4.0\, s$:$4.0 = 0 + \frac{1}{2} a (4.0)^2 \Rightarrow 4.0 = 8a \Rightarrow a = 0.5\, m/s^2$.$4$. Substituting $a$ into the force equation (taking $g = 10\, m/s^2$):$0.5 = 10 \sin 30^o - \mu_k (10) \cos 30^o$$0.5 = 10(0.5) - \mu_k (10)(\frac{\sqrt{3}}{2})$$0.5 = 5 - 5\sqrt{3} \mu_k$$5\sqrt{3} \mu_k = 4.5$$\mu_k = \frac{4.5}{5 	imes 1.732} = \frac{0.9}{1.732} \approx 0.519 \approx 0.5$.Thus,$\mu_s \approx 0.6$ and $\mu_k \approx 0.5$.

Components	$\%$ of the total cellular mass
$A$. Water	$1$. $70-90$
$B$. Protein	$2$. $10-15$
$C$. Carbohydrates	$3$. $3$
$D$. Lipids	$4$. $2$
$E$. Nucleic acids	$5$. $5-7$
$F$. Ions	$6$. $1$

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