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3 years ago

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6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by

an unbalanced force. How do these experiments show this?

2 answers:

Nitella [24]3 years ago

7 0

I’m not sure what experiments you’re referring to. Maybe if there is a Frictional force that stops and object from sliding after a period of time instead of it continuing to slide forever.

gavmur [86]3 years ago

5 0

You can test if it’s true by holding a pencil in mid air over a table and the table is supposed to be the unbalanced forced that stopped the pencil from moving at the constant velocity it was going by.

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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 plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F

vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

$t = \frac{d}{v_{net}}$

here net speed of the plane will be given as

$v_{net} = v + v_w$

$t = \frac{1000}{205 + v_w}$

similarly when it moves under the condition of headwind its net speed is given as

$v_{net} = v - v_w$

now time taken to cover the distance is 2 hours more

$t + 2 = \frac{1000}{205 - v_w}$

now solving two equations

$\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}$

solving above for v_w we got

$v_w = 40.4 mph$

6 0

3 years ago

Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change

Ket [755]

Answer:

a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

Explanation:

Let consider the bottom of the first floor in a building as the zero reference ( $z = 0\,m$ ). The change in potential energy experimented by a particle ( $\Delta U_{g}$ ), measured in joules, is:

$\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$z_{o}$ , $z_{f}$ - Initial and final height with respect to zero reference, measured in meters.

Please notice that $m\cdot g$ is the weight of the particle, measured in newtons.

a) If we know that $m = 3\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{o} = 0\,m$ and $z_{f} = 4.1\,m$ , then the change in potential energy is:

$\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)$

$\Delta U_{g} = 120.626\,J$

The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) If we know that $m\cdot g = 650\,N$ , $z_{o} = 4.1\,m$ and $z_{f} = 0\,m$ , then the change in potential energy is:

$\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)$

$\Delta U_{g} = -2665\,J$

The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

5 0

3 years ago

What is the effect of the mass of an object on it’s friction threshold?

Alina [70]

An object of large mass is pulled down onto a surface with a greater force than an object of low mass and, as a consequence, there is greater friction between the surface of the heavy object than between the surface and the light object.

6 0

3 years ago

Help please, I don't get it

Sergio [31]

Answers:

a) $\hat F=(0.83,-0.55) N$

b) $\hat D=(-0.44,-0.89) m$

c) $\hat V=(-0.47,0.88) m/s$

Explanation:

A unit vector is a vector whose magnitude (length) is equal to 1. This kind of vector is identified as $\hat v$ and the way to calculate is as follows:

$\hat v=\frac{\vec v}{|v|}$

Where:

$\vec v=(x,y)$ is the vector

$|v|=\sqrt{x^{2}+y^{2}}$ is the magnitude of the vector

Having this information clarified, let's begin with the answers:

a) Force Vector

$\vec F=(9.0 \hat i - 6.0 \hat j) N$

Magnitude of $\vec F$ :

$|F|=\sqrt{(9.0 \hat i)^{2}+(-6.0 \hat j)^{2 }}N=10.81 N$

<u />

<u>Unit vector:</u>

$\hat F=\frac{\vec F}{|F|}$

$\hat F=\frac{(9.0 \hat i - 6.0 \hat j) N}{10.81 N}$

$\hat F=\frac{9.0}{10.81} N-\frac{6.0}{10.81}N$

$\hat F=(0.83,-0.55) N$

b) Displacement Vector

$\vec D=(-4.0 \hat i - 8.0 \hat j) m$

Magnitude of $\vec D$ :

$|D|=\sqrt{(-4.0 \hat i)^{2}+(-8.0 \hat j)^{2 }}m=8.94 m$

<u />

<u>Unit vector:</u>

$\hat D=\frac{\vec D}{|D|}$

$\hat D=\frac{(-4.0 \hat i - 8.0 \hat j) m}{8.94 m}$

$\hat D=\frac{-4.0}{8.94} Nm+\frac{-8.0}{8.94}m$

$\hat D=(-0.44,-0.89) m$

c) Velocity Vector

$\vec V=(-3.50 \hat i + 6.50 \hat j) m/s$

Magnitude of $\vec V$ :

$|V|=\sqrt{(-3.50 \hat i)^{2}+(6.50 \hat j)^{2}}m/s=7.38 m/s$

<u />

<u>Unit vector:</u>

$\hat V=\frac{\vec V}{|V|}$

$\hat V=\frac{(-3.50 \hat i +6.50 \hat j) m/s}{7.38 m/s}$

$\hat V=\frac{-3.50}{7.38} m/s+\frac{6.50}{7.38}m/s$

$\hat V=(-0.47,0.88) m/s$

8 0

4 years ago