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serg [7]
2 years ago
14

If a ball rolling down a hill is half way between the top and bottom, how much potential energy does the ball have compared to k

inetic energy?
Chemistry
1 answer:
madreJ [45]2 years ago
3 0

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength g is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If m denote the mass of this ball, the gravitational potential energy (\rm GPE) of this ball at height h would be {\rm GPE} = (m \cdot g) \cdot h, which is proportional to h\!.

The value of g near the surface of the earth is indeed approximately constant (typically g \approx 9.8\; \rm m \cdot s^{-2}.)

At halfway between the top and bottom of this hill, the height of this ball would be (1/2) of its initial value (the value when the ball was at the top of the hill.) Because the \rm GPE of this ball is proportional to its height, at halfway down the hill, the \rm GPE\! of this ball would also be (1/2)\! its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that (1/2) of the initial \rm GPE\! of this ball was not lost. Rather, these (1/2)\! of the initial \rm GPE would have been converted to the kinetic energy (\rm KE) of this ball.

Hence, when the ball is halfway down the hill:

\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}.

\begin{aligned}& \text{KE halfway down the hill}\\  &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}.

Therefore:

\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}.

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

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Can Magnesium hydroxide be used as the base for this titration? Why or why not? 5. Describe the procedure that you will follow t
kvasek [131]

Explanation:

A.

Yes

B.

2KHP + Mg(OH)2 = 2KOH + Mg(HP)2

KHP is an acidic salt which reacts with Mg(OH)2 which is a base via double displacement/decomposition reaction.

C.

• One approach is to prepare the solution volumetrically using KHP crystals.

• Preparing 1 liter of 0.1 M KHP you would add 0.1 moles of KHP

Molar mass of KHP(C8H5KO4)

= (12*8) + (5*1) + 39 + (16*4)

= 204 g/mol

Number of moles = mass/molar mass

= 20.4/204

= 0.1 mol.

• This is added to a 1 liter volumetric flask, add deionized water until near the fill line, stopper and mix

D.

Number of moles = molar concentration * volume

= 0.025 * 0.100

= 0.0025 mol

Equation for the reaction:

Ba(OH)2 + 2KHP --> Ba(KP)2 + 2H2O

1 mole of Ba(OH)2 reacted with 2 moles of KHP. By stoichiometry,

Number of moles of Ba(OH)2 = 0.0025/2

= 0.00125 mol.

Molarity = number of moles/volume

= 0.00125/0.01

= 0.125 M of Ba(OH)2.

4 0
3 years ago
What would be the mass in grams of 1.204 x 1024 molecules of sulfur dioxide
melomori [17]

Answer:

mass of sulfur = 96 g

Explanation:

no of moles of sulfur dioxide in 1.204\times 10^{24} molecules = \frac{1.204\times 10^{24}}{avagadro number }= \frac{1.204\times 10^{24}}{6.023\times 10^{23}}

                                       = 2 moles

therefore mass of sulfur dioxide = moles×atomic number

                                                      =2×(16+32)

                                                      =96

6 0
3 years ago
Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh
Aliun [14]

Answer:

1.60.

Explanation:

  • The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
  • The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>

<em></em>

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>

5 0
3 years ago
Calcule la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) que se preparó al mezclar, en un recipiente aforado, 4 mo
oee [108]

Considerando la definición de molaridad, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5 \frac{moles}{litro}.

La molaridad es una medida de la concentración de un soluto en una disolución que se define como el número de moles de soluto que están disueltos en un determinado volumen.

La molaridad de una solución se calcula dividiendo los moles del soluto por el volumen de la solución:

Molaridad=\frac{numero de moles de soluto}{volumen}

La Molaridad se expresa en las unidades \frac{moles}{litro}.

En este caso, sabes que una solución acuosa se preparó al mezclar 4 moles del ácido con suficiente agua hasta completar 8 litros de solución. Entonces, sabes que:

  • número de moles de soluto= 4 moles
  • volumen= 8 litros

Reemplazando en la definición de molaridad:

Molaridad=\frac{4 moles}{8 litros}

Resolviendo:

Molaridad= 0.5 \frac{moles}{litro}

Finalmente, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5 \frac{moles}{litro}.

<em>Aprende más</em>:

  • brainly.com/question/17647411?referrer=searchResults
  • brainly.com/question/21276846?referrer=searchResults

8 0
2 years ago
Draw a bond-line structure for CH3CH2O(CH2)2CH(CH3)2.<br><br> Include Lone Pairs in your answer.
mash [69]

Answer:

See explanation and image attached

Explanation:

A bond line structure refers to any structure of a covalent molecule wherein the covalent bonds present in the molecule are represented with a single line for each level of bond order.

The bond-line structure of CH3CH2O(CH2)2CH(CH3)2 has been shown in the image attached. We know that oxygen has a lone pair of electrons and this has been clearly shown also in the image attached.

7 0
3 years ago
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