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Gennadij [26K]
3 years ago
5

Un atleta se dirige hacia la línea de salida a 6 m/s. Cuando pasa por la marca de 100 m, ponemos el cronómetro en marcha. Escrib

e las ecuaciones del movimiento re represéntalas gráficamente.
Chemistry
1 answer:
Romashka [77]3 years ago
4 0

Answer:

Los parámetros dados del movimiento son;

La velocidad con la que el atleta se dirige hacia la línea de salida = 6 m / s

El punto en el que se inicia el cronómetro = marca de 100 m

La ecuación para la distancia recorrida, 's', se da como sigue;

s = 6 m / s × t

La distancia total recorrida, d = 100 + s

∴ d = 100 + 6·t

Los gráficos de las ecuaciones se crean con Microsoft Excel y se presentan de la siguiente manera;

Explanation:

You might be interested in
Neon atom can exist freely in mature but sodium atom can not,why
andrew-mc [135]

Answer:-

Neon is a noble gas. Neon with an electronic configuration of 2,8 has 8 electrons in it's outermost shell or valence shell.

According to the Octet theory if an element has eight electrons in it's valence shell then it is stable and does not undergo reaction. Thus Neon does not need to react and can exist freely in nature.

Sodium with an electronic configuration of 2,8,1 has 1 electron in it's valence shell. As per octet rule, it is not stable and it must lose that 1 electron to become stable.

In order to lose that 1 electron sodium must react with other substances. Hence sodium cannot exist freely in nature.

5 0
3 years ago
Two liters of a perfect gas are at 0>c and 1 atm. if the gas is nitrogen, n2, determine the number of molecules.
ZanzabumX [31]

Solution;

The gas is at STP;

Where; T = 273 K , P = 1 atm

We know that 1 mole of a gas at STP occupies a volume of 22.4 liters .

V1/n1 = V2/n2

n2 = (V2/V1) n1

n2 = (2 l/22.4 l)(1 mole)

n2 = 0.0893 moles

But 1 mole of a compound has 6.022 × 10^23 molecules

Thus, number of molecules = 0.0893 moles × 6.022 ×10^23 molecules

= 5.378 × 10^22 molecules.

4 0
4 years ago
A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each
ale4655 [162]

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.  

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 m_3. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to m_3 /day. In that sense, we can calculate the retention time using the tank volume in m_3.

Ft =12 \frac{MG}{day}  (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}

After that, we should divide the total flow by four, because we have four tanks.

F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}

To calculate the retention time we divide the total volume V by the flow of each tank F1.

t1 =\frac{V1}{F1} = \frac{600 m^3}{11340  \frac{m^3}{d} }  = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

b)

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

t2 =\frac{V1}{F2} = \frac{600 m^3}{45360  \frac{m^3}{d} }  = 0.01 day\\t2 = 0.01  d (\frac{24 h}{1d} ) = 0.31 h

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

5 0
3 years ago
Activation energy is _____.
Archy [21]
<span>Activation energy is _____
</span>
<span>an energy barrier between reactants and products
</span>
7 0
3 years ago
Read 2 more answers
What is the volume of 0.80 grams of O2 gas at STP?
frez [133]

Answer:

The volume of 0.80 grams of O2 is 0.56 L

Explanation:

Step 1 : Data given

Mass of O2 = 0.80 grams

Molar mass of O2 = 32 g/mol

STP = 1 mol, 1atm, 22.4 L

Step 2: Calculate moles of oxygen

Moles of O2 = Mass of O2 / molar mass of O2

Moles O2 = 0.80 grams / 32 g/mol

Moles O2 = 0.025 moles

Step 3: Calculate volume of O2

1 mol = 22.4 L of gas

0.025 moles = 0.025*22.4L = 0.56 L

The volume of 0.80 grams of O2 is 0.56 L

5 0
4 years ago
Read 2 more answers
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