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2 years ago

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A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

e block take to reach the bottom?

1 answer:

Snowcat [4.5K]2 years ago

4 0

Answer:

$\boxed{\text{\sf \Large 3.0 s}}$

Explanation:

Use distance formula

$\displaystyle d=ut+\frac{1}{2} at^2$

$u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}$

$\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2$

$t=3$

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The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

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When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

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When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

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The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

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The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

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(Rounded to $\text{$3$ sig. fig.}$ )

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