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yulyashka [42]
3 years ago
13

Please help!Much appreciated!​

Physics
1 answer:
lutik1710 [3]3 years ago
4 0

Answer:

Ts = 3.0×10¯² sec

Explanation:

From the question given:

Ts = 2π √(4.5×10¯² Kg / 2.0×10³ Kg/s²)

The above expression can be evaluated as follow:

Pi (π) = 3.14

Ts = 2π √(4.5×10¯² / 2.0×10³)

Ts = 2 × 3.14 √(4.5×10¯² / 2.0×10³)

Ts = 6.28 × √(2.25×10¯⁵)

Ts = 6.28 × 4.74×10¯³

Ts = 3.0×10¯² sec

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kykrilka [37]

The specific heat capacity of the substance is 0.963 J/g^{\circ}C

Explanation:

When an object of mass m is supplied with a certain amount of energy Q, its temperature increases according to the equation:

Q=mC_s \Delta T

where

m is the mass of the object

C_s is its specific heat capacity

\Delta T is the increase in temperature of the object

In this problem, we have

Q=300 cal \cdot 4.814 = 1444.2 J

m = 50 g

\Delta T = 20C-(-10C)=30^{\circ}C

Therefore, we can solve for C_s to find its specific heat capacity:

C_s = \frac{Q}{m\Delta T}=\frac{1444.2}{(50)(30)}=0.963 J/gC

Learn more about specific heat capacity:

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4 years ago
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kotegsom [21]

Answer:

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In this problem, the object is moved at a distance twice as far as the initial point, so

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3 years ago
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frosja888 [35]

Answer

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s2008m [1.1K]

Answer:

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Explanation:

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La excentricidad e de una elipse es una medida de lo alejado que se encuentran los focos del centro.

Pues bien, la mayoría de las órbitas planetarias tienen un valor muy pequeño de excentricidad, es decir e ≈ 0. Esto significa que, a nivel práctico, pueden considerarse círculos descentrados.

Segunda Ley:

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Tercera ley de Kepler:

La tercera ley, también conocida como armónica o de los periodos, relaciona los periodos de los planetas, es decir, lo que tardan en completar una vuelta alrededor del Sol, con sus radios medios.

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3 years ago
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