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3 years ago

14

A car speeds up from 14 meters per second to 21 meters per second in 6 seconds. Whats the acceleration and the distance passed w

ithin that time?

1 answer:

Solnce55 [7]3 years ago

7 0

Answer:

a = 1.666... m/s²

Explanation:

a = v2 - v1 / t2 - t1

a = 21m/s - 14m/s / 6s - 0s

a = 7m/s / 6s

a = 1.666... m/s²

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Percent Yield Lab Report

Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0

3 years ago

A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (

Margarita [4]

$\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1/4 \ mile = 402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}$

4 0

2 years ago

a water pump takes water through 12m. Water is discharged at the surface at the rate of 2Kg/s, and the discharge speed of each d

pickupchik [31]

Answer:

45

Explanation:

3 0

3 years ago

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.

iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g

8 0

4 years ago

A listener is sitting somewhere on the line between two loudspeakers that are 10 m apart. The speakers are each emitting a sine

prohojiy [21]

Answer:

57.17 Hz 114.34 Hz 285 Hz

Explanation:

The distance between the men and 1 speaker = 3.5 m

Distance between the men and second speaker = 10-3.5= 6.5 m

Here at this point there will be no sound so there will be destructive interference

Path difference $\Delta x=6.5-3.5=3$

We know that for destructive interference $\Delta x=(2m+1)\frac{\lambda }{2}=(2m+1)\frac{v}{2f}$

$3=(2m+1)\frac{v}{2f}$

$f=(2m+1)\frac{v}{6}$ here v is the speed of sound in air

So for m =0

$f=(2\times 0+1)\frac{343}{6}=57.17H$

for m =1

$f=(2\times 1+1)\frac{343}{6}=114.34Hz$

for m=2

$f=(2\times 2+1)\frac{343}{6}=285Hz$

6 0

3 years ago