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3 years ago

Work and power!!

Physics

1 answer:

Gnesinka [82]3 years ago

6 0

Answer:

Efficiency

Explanation:

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The volume of a gas decreases from 15.7 mºto 11.2 m3 while the pressure changes from 1.12 atm to 1.67 atm. If the

Svetlanka [38]

Answer:

Approximately $261\; \rm K$ , if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let $P_1$ and $P_2$ denote the pressure of this gas before and after the changes.

Let $V_1$ and $V_2$ denote the volume of this gas before and after the changes.

Let $T_1$ and $T_2$ denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let $n_1$ and $n_2$ denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios $\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}$ and $\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}$ should both be equal to the ideal gas constant, $R$ .

In other words:

$R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}$ .

$R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}$ .

Combine the two equations (equate the right-hand side) to obtain:

$\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}$ .

Rearrange this equation for an expression for $T_2$ , the temperature of this gas after the changes:

$\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1$ .

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: $n_2 = n_1$ , $(n_2 / n_1) = 1$ .

$\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}$ .

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3 years ago

Which gas in Earth's atmosphere has increased over time due to burning fossil fuels?

astraxan [27]

The answer is Carbon Dioxide

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3 years ago

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An object at rest or moving at a constant speed in a straight path will continue to do so unless acted upon by an unbalanced for

Furkat [3]

I believe if your looking for true or false answer, that the answer is true

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3 years ago

The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus)

Artemon [7]

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge $q=5\sin4\pi t\ mC$

Voltage $v=3\cos4\pi t\ V$

Time t = 0.3 sec

We need to calculate the current

Using formula of current

$i(t)=\dfrac{dq}{dt}$

Put the value of charge

$i(t)=\dfrac{d}{dt}(5\sin4\pi t)$

$i(t)=5\times4\pi\cos4\pi t$

$i(t)=20\pi\cos4\pi t$

(a).We need to calculate the power delivered to the element

Using formula of power

$p(t)=v(t)\times i(t)$

Put the value into the formula

$p(t)=3\cos4\pi t\times20\pi\cos4\pi t$

$p(t)=60\pi\times10^{-3}\cos^2(4\pi t)$

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})$

Put the value of t

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})$

$p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)$

$p(t)=187.68\ mW$

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

$E(t)=\int_{0}^{t}{p(t)dt}$

Put the value into the formula

$E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}$

$E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}$

$E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}$

$E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)$

$E(t)=57.52\ mJ$

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

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3 years ago

What are some of the major differences between the ancient Olympics and modern-day Olympics? List and describe at least two diff

umka21 [38]

Answer:

The ancient Olympic games only allowed people of Greek descent to participate. The Salt Lake City Olympics featured 2600 athletes from 77 countries. Only a few hundred athletes participated in the ancient games.

Only men were allowed to compete in the ancient Greek games. Athletic training in ancient Greece was part of every free male citizen's education. The first women to compete in the Olympics were Marie Ohnier and Mme. Brohy. They participated in croquet games in the 1900 Olympics.

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3 years ago