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oksano4ka [1.4K]

4 years ago

13

While doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm3. The accepted value for the density

of aluminum is 2.70 g/cm3. What was the student's percent error?

1 answer:

bezimeni [28]4 years ago

3 0

The equation for percent error is

% Error = $100*|Experimental-Theoretical|/Theoretical$

Our experimental is 2.85g/cm^3 and the accepted is 2.7g/cm^3

Thus our % Error = 5.555%

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4) A satellite, mass m, is in circular orbit (radius r) around the earth, which has mass ME and radius Re. The value of r is lar

defon

<h2>Answers:</h2>

(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:

$K=\frac{1}{2}m{V}{2}$ (1)

Where $m$ is the mass of the body and $V$ its velocity.

In this specific case of the satellite, its kinetic energy $K_m$ taking into account its mass $m$ is:

$K_{m}=\frac{1}{2}m{V}^{2}$ (2)

On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:

$V=\sqrt{G\frac{ME}{r}}$ (3)

Where $G$ is the gravity constant, $ME$ the mass of the Earth and $r$ the radius of the orbit <u>(measured from the center of the Earth to the satellite). </u>

Now, if we substitute the value of $V$ from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:

$K_{m}=\frac{1}{2}m{\sqrt{G\frac{ME}{r}}}^{2}$ (4)

Finally:

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5) >>>>This is the kinetic energy of the satellite

(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period $T$ is defined as:

$T=2\pi\sqrt{\frac{r^{3}}{\mu}}$ (6)

Where $\mu$ is a constant and is equal to $GME$ . So, this equation in these terms is written as:

$T=2\pi\sqrt{\frac{r^{3}}{GME}}$ (7)

As we can see, <u>the Period of the orbit does not depend on the mass of the satellite </u> $m$ , it depends on the mass of the greater body (the Earth in this case) $ME$ , the radius of the orbit and the gravity constant.

(c) The gravitational force described by the law of gravity is a central force and therefore is <u>a conservative force</u>. This means:

1. The work performed by a gravitational force to move a body from a position A to a position B <u>only depends on these positions and not on the path followed to get from A to B. </u>

2. When the path that the body follows between A and B is a c<u>losed path or cycle</u> (as this case with a <u>circular orbit</u>), <u>the gravitational work is null or zero</u>.

<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, <u>it always acts perpendicular to the movement</u>. </h2>

Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.

(d) The total Mechanical Energy $E$ of a body is the sum of its Kinetic Energy $K$ and its Potential Energy $P$ :

$E=K+P$ (8)

But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5)

And its potential energy due to the Earth gravitational field as:

$P_{m}=-G\frac{mME}{r}$ (9)

This energy is negative by definition.

So, the total mechanical energy of the orbit, also called the Orbital Energy is:

$E=\frac{1}{2}Gm\frac{ME}{r}+(- G\frac{mME}{r})$ (10)

Solving equation (10) we finally have the Orbital Energy:

$E=-\frac{1}{2}mME\frac{G}{r}$ (11)

At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy $E$ :

-When $E=0$ :

We are talking about an <u>open orbit</u> in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:

$K_{m}=-P_{m}$

Such is the case of some comets in the solar system.

-When $E>0$ :

We are also talking about <u>open orbits</u>, which are hyperbolic, being $K_{m}>P_{m}$

<h2>-When $E$ : >>>><u>This case</u></h2>

We are talking about <u>closed orbits</u>, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:

-Elliptical orbit: Although $E$ is constant, $K_m$ and $P_m$ are changing along the trajectory .

-Circular orbit: When at all times both the kinetic energy $K_m$ and the potential $P_m$ remain constant, resulting in a total mechanical energy $E$ as the one obtained in this exercise. This means that the speed is constant too and <u>is the explanation of why this Energy has a negative sign. </u>

3 0

4 years ago

Find the mass of a person who weighs 608 N

Anvisha [2.4K]

Answer:

$weight = mass \times accelaration \: due \: to \: gravity \\ 608 = x \times 10 \\ x = 608 \div 10 = 60.8 \\ mass = 60.8 \\ thank \: you$

7 0

3 years ago

Read 2 more answers

Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas

Burka [1]

Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = $9.232 g/cm^3$

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}$

$a = 3.35\times 10^{-8} cm$

Atomic radius of the polonium in unit cell = r

r = 0.5a

$r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm$

$1 cm = 10^{10} pm$

$1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm$

Approximate atomic radius for polonium-209 is 167.5 pm.

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3 years ago

The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

bagirrra123 [75]

The velocity of the wave on the string is given by

$v=\sqrt{\frac{T}{\frac{m}{L}}} \\ v=\sqrt{\frac{TL}{m}}$

Solving the above equation,

$v^2=\frac{TL}{m} \\ L=\frac{v^2m}{T}$

The frequency of the wave $f=300$ and wave length is $0.76$

The velocity is $v=(300)(0.76)=228$

Substituting numerical values,

$L=\frac{228^2(0.0059)}{200}\\ T=1.534$

The length of the string is $1.534 m$

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3 years ago

The dimensions of aluminum foil in a box for sale in super markets are 66 2/3 yards by 12 inches. the mass of the foil is 0.83 k

disa [49]

Length of the sheet is given as

$L = \frac{200}{3} yards = 6096 cm$

width of the sheet is given as

$w = 12 inches = 30.48 cm$

now let say its thickness is "t"

so the volume of the sheet is given as

$V = L*w*t$

$V = 6096*30.48* t$

$V = 185806.08*t cm^3$

mass of the sheet is given as

$m = 0.83 kg = 830 gram$

now we have

$density = \frac{mass}{volume}$

$2.70 = \frac{830}{185806.08*t}$

by solving above we have

$t = 1.65 * 10^{-3} cm$

so the thickness of sheet will be above

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3 years ago