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2 years ago

A gas is confined in a cylinder with a piston. What happens when work is done on the gas.

Physics

2 answers:

goblinko [34]2 years ago

5 0

It combusts in to a vapor

yanalaym [24]2 years ago

3 0

The correct answer is:

The volume of the gas decreases

Explanation:

As the piston is forced down work is done on the gas. The force exercised by the gas pressure on the piston is in the adverse direction of the motion of the piston. The volume of the space the gas absorbs decreases. When we let go of the piston the gas extends. The force exercised by the gas is in the similar direction as the motion of the piston and work is done by the system.

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Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co

patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of $^{3890}Sr\, t_{\frac{1}{2}}$ = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

$\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where

N = No. of nuclei left after time t

$N_{o}$ = No. of nuclei initially started with

$\frac{N}{N_{o}} = 1\times 10^{- 4}$

(Since, 100% - 99.99% = 0.01%)

Thus

$1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}$

Taking log on both the sides:

$- 4 = \frac{t}{28.5}log\frac{1}{2}$

$t = \frac{-4\times 28.5}{log\frac{1}{2}}$

t = 387.69 yrs

5 0

3 years ago

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables

Juli2301 [7.4K]

They made me do it I don’t even know what to say I’m so sorry

7 0

3 years ago

Give one example of a thermodynamic cycle that does not account for the carnot efficiency.

Arturiano [62]

Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.

The Carnot cycle is a hypothetical cycle that takes no account of entropy generation. It is assumed that the heat source and heat sink have perfect heat transfer. The working fluid also remains in the same phase, as opposed to the Rankine cycle, in which the fluid changes phase. A practical thermodynamic cycle, such as the Rankine cycle, would achieve at most 50% of the Carnot cycle efficiency under similar heat source and heat sink temperatures.

<h3>What is Thermo-Electrochemical converter?</h3>

In a two-cell structure, a thermo-electrochemical converter converts potential energy difference during hydrogen oxidation and reduction to heat energy.

It employs the Ericsson cycle, which is less efficient than the Carnot cycle. In a closed system, it converts heat to electrical energy. There are no external input or output devices.

This means there will be no mechanical work to be done, as well as no exhaust. As a result, Carnot efficiency is not taken into account in this cycle. Carnot efficiency is accounted for by other options such as turbine and engine.

Learn more about Thermo-Electrochemical converter here:

brainly.com/question/13040188

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4 0

1 year ago

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed o

adell [148]

Answer:

Part a)

$v_f = 7.99 m/s$

Part b)

$y = 3.25 m$

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

$v_f^2 - v_i^2 = 2 a y$

here we have

$v_i = 2.22 m/s$

$y = -3 m$

$v_f^2 - (2.22)^2 = 2(-9.81)(-3)$

$v_f = 7.99 m/s$

Part b)

at highest point of his motion the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a (\Delta y)$

$0 - 2.22^2 = 2(-9.81)(y - 3)$

$y = 3.25 m$

7 0

3 years ago

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human i

Varvara68 [4.7K]

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$ .

7 0

2 years ago