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2 years ago

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Which of the following is/are true? If caught in a rip current, swim directly towards shore, using the most powerful kick possib

le. If caught in a rip current, swim perpendicular to shore, until you are completely out of the rip current, then swim in towards shore. If caught in a rip current, swim parallel to shore, until you are completely out of the rip current, then swim in towards shore. Don't fight the rip current; relax and let it carry you closer to shore.

1 answer:

In-s [12.5K]2 years ago

4 0

hi true false true false

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Which of the following is a chemical equation that accurately represents what happens when sulfur and oxygen react to form sulfu

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<span>D. 2S + 3O2 right-pointing arrow 2SO3 Let's look at the 4 options and see what the issues are with each of the wrong ones. A. Sulfur and oxygen react to form sulfur trioxide. * This is just a quick description, but nothing quantitative. How much sulfur? How much oxygen? Are the reactants single atoms or molecules? Not enough information, so this is the wrong choice. B. S and O2 produce SO3. * Just replacing words with element abbreviations does't fix the problem. So it's still wrong. C. S + O2 right-pointing arrow SO3 * This is getting closer. We have the correct symbols for the reactants and product. But the equation doesn't balance. On the left hand side, there are 2 oxygen atoms and on the right hand side, there's 3. Where did the 3rd oxygen atom come from? So it's still wrong. D. 2S + 3O2 right-pointing arrow 2SO3 * This is correct. We know that we're using elemental sulfur and an oxygen molecule on the left hand side. Actually 2 atoms of elemental sulfur and 3 molecules of oxygen gas. And we're producing 2 molecules of sulfur trioxide. And the numbers of each type of atom match in count on both sides. So this is the correct answer.</span>

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3 years ago

"a 1,600 kg car is traveling at a speed of 12.5 m/s. what is the kinetic energy of the car?"

xxTIMURxx [149]

I had this question before can you show me the answers choices

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2 years ago

Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of

notsponge [240]

Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force ( $F$ ), measured in newtons, is the product of mass ( $m$ ), measured in kilograms, and net acceleration ( $a$ ), measured in meters per square second. That is:

$F = m\cdot a$ (1)

The initial force applied in the mass is:

$F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)$

$F = 25\,N$

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to $\frac{4}{5}$ of the initial force. The acceleration of the mass is:

$\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}$

$a = 2\,\frac{m}{s^{2}}$

The acceleration of the mass is 2 meters per square second.

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2 years ago

the field of meteorology is based on studying which of the following? a.)stars and plants b.) weather and climate c.)rocks and m

ipn [44]

B) Weather and Climate

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2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago