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VLD [36.1K]
3 years ago
6

Which statement was part of an initial draft of the Declaration of Independence, but not part of the final version of the Declar

ation of Independence?
A. Britain took away natural rights of colonists.
B. Slavery should be outlawed in the new country.
C. The United States of America declared itself independent.
D. Governments were created by people.
Physics
1 answer:
aleksandrvk [35]3 years ago
3 0
The correct option is B.
The idea that slavery should be abolished in America was part of the initial draft of the Declaration of independence but it was not included as part of the final version of the document, because some Americans were of the opinion that such move might hurt the economy of the country.
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How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position
e-lub [12.9K]
The answer is C. Final position minus initial position.
5 0
3 years ago
Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
What is the displacement of the runner in 4 laps? *<br> 2 point
Serhud [2]

Answer:

There is no displacement.

Explanation:

Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.

Hope this helps. . .

ly UwU

3 0
2 years ago
A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How
Sati [7]

0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>
  • Mathematically, the work done to strech a string is given as 1/2 ×K×x².
  • K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
  • The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
  • Then K = Force / x.
  • Here x = 15 - 10 = 5 cm = 0.05 m
  • K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
  • Work done = 1/2×k×x²
  • Here x= 18-15=3cm or 0.03 m
  • So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

#SPJ4

6 0
1 year ago
Calculate the force it would take to accelerate a 50 kg bike at a rate of 3 m/s2.
Ede4ka [16]

ummmm it might be 300... i used a calculator

sorry if it is wrong

4 0
3 years ago
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