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4 years ago

A rock bounces down a mountain. When will its kinetic energy be greatest?

Chemistry

2 answers:

Anuta_ua [19.1K]4 years ago

8 0

When the Rock is about to hit the ground

dedylja [7]4 years ago

4 0

Answer:

The rock gains its maximum amount of K.E before striking the ground surface

Explanation:

When an object is comprised of certain energy due to its relative height, then it is known as the potential energy (P.E) and when an object is composed of energy due to its motion then its called the kinetic energy (K.E). These two energies are associated with one another.

An object when placed at a certain height, it contains the maximum amount of P.E. and absence of K.E. As the object is allowed to move downward, the P.E starts decreasing and the K.E starts increasing. Before the point of contact (collision) between the object and the ground surface, there will be maximum K.E and no P.E in the object. After it hits the ground, this energy gets transfer into the sound energy.

Hence, it can be concluded that when a rock bounces down a mountain, it gains its maximum amount of K.E before striking the ground surface.

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If a 520 mg sample of technetium-99 is used for diagnostic procedure, how much of Tc-99 remains after 30.0h? Half life of Tc-99

k0ka [10]

16.25 mg
I have attached an image of the half life formula.
In this problem, N0 would be 520, since that is the initial amount.
t = 30.0 Hours
t 1/2 = 6.0 hours
Now plug everything into the equation. You get 16.25 mg. No

4 0

3 years ago

How many grams of Br2 are needed to form 67.1 g of AlBr3 ?<br><br> 2Al(s)+3Br2(l)⟶2AlBr3(s)

Romashka [77]

Answer:

Approximately $60.3\; \rm g$ .

Explanation:

Look up the relative atomic mass of $\rm Al$ and $\rm Br$ on a modern periodic table:

$\rm Al$ : $26.982$ .
$\rm Br$ : $79.904$ .

Calculate the formula mass of $\rm AlBr_3$ and $\rm Br_2$ :

$\begin{aligned}& M(\mathrm{AlBr_3}) = 26.982 + 3 \times 79.904 \approx 266.694\; \rm g \cdot mol^{-1} \\ & M(\mathrm{Br_2}) = 2\times 79.904 \approx 159.808\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $67.1\; \rm g$ of $\rm AlBr_3$ :

$\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{m(\mathrm{AlBr_3})}{M(\mathrm{AlBr_3})} \\ &\approx \frac{67.1\; \rm g}{266.694\; \rm g \cdot mol^{-1}} \approx 0.2516\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The ratio between the coefficients of $\rm Br_2$ and $\rm AlBr_3$ in that equation is three-to-two. That corresponds to the ratio:

$\begin{aligned}\frac{n(\text{$\mathrm{Br_2}$, consumed})}{n(\text{$\mathrm{AlBr_3}$, produced})} &= \frac{3}{2}\end{aligned}$ .

It is already calculated that approximately $0.2516\; \rm mol$ of $\rm AlBr_3$ was produced through this reaction. Apply this ratio to approximate the (minimum) number of moles of $\rm Br_2$ that is consumed:

$\displaystyle \frac{3}{2} \times 0.2516\; \rm mol \approx 0.3774\; \rm mol$ .

Calculate the mass of that $0.3774\; \rm mol$ of $\rm Br_2$ :

$\begin{aligned}m(\mathrm{Br_2}) &= n(\mathrm{Br_2})\cdot M(\mathrm{Br_2}) \\ &\approx 0.3774\; \rm mol \times 159.808\; \rm g \cdot mol^{-1} \approx 60.3\; \rm g\end{aligned}$ .

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3 years ago

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MakcuM [25]

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3 years ago

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Phoenix [80]

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4 years ago

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