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noname
9 days ago
15

In pottery class, you throw a pot from a lump of wet clay. your pot's mass is 5.5 kg. after the pot is fired, it's mass is 4.9 k

g. the density of wet clay is 1.60 g/cm3 and the density of fired clay is 1.36 g/cm3. what was the volume of your pot before it was fired? what was the volume of the pot after it was fired? express your answer in significant digits.
Physics
1 answer:
ValentinkaMS [3K]9 days ago
8 0

To find the volume, we can utilize the ratio of mass to density, as shown by:

volume = mass / density

 

A. when mass = 5.5 kg = 5500 g; density = 1.60 g/cm^3

volume = 5500 g / (1.60 g/cm^3)

resulting in volume = 3,437.5 cm^3

By rounding according to significant digits:

volume = 3,400 cm^3 = 3.4 L

 

B. when mass = 4.9 kg = 4900 g; density = 1.36 g/cm^3

volume = 4900 g / (1.36 g/cm^3)

calculating gives volume = 3,602.94 cm^3

Considering significant digits:

<span>volume = 3,600 cm^3 = 3.6 L</span>

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Answer:

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d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

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n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrones.

b) La densidad de corriente 'J' en el cable se calcula como

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J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) La velocidad típica 'V_{d' de un electrón se expresa como:

V_{d = \frac{J}{n|q|}

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V_{d =1.114 x 10^{4m/s

d) De acuerdo con estas ecuaciones,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

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1 month ago
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Answer:

0.0984

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Substituting 42° for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

Meanwhile, the vertical component is:

F_Y = Fsin \ \theta

Again substituting 42° for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

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In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

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Where;

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Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting F_Y \ for \ F_{sin \ \theta}. Normal reaction becomes:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

F_X = \mu \ ( mg - \ F_Y)

Rearranging the equation above to isolate \mu leads to:

\mu \ = \ \frac{F_X}{mg - F_Y}

Substituting in the following values:

F_X \ = \ 64.65 \ N

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Thus:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

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Answer:

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