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3 years ago

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A sample of an ideal gas initially occupies a volume of 6 L. The pressure of the sample is then doubled while it is cooled to on

e third its original absolute temperature. After these changes, what volume does the sample occupy?

1 answer:

Troyanec [42]3 years ago

4 0

Answer:

V₂= 1 L

Explanation:

Given that

Volume occupies V₁= 6 L

Initial pressure = P₁

Initial temperature = T₁

The final pressure =P₂ = 2 P₁

Final volume =V₂

Final temperature = T₁/3

As we know that equation for ideal gas

P V = m R T

P=pressure, V=volume, T=temperature

m=mass ,R=gas constant

Now from mass conservation

$m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}$

$\dfrac{P_1\times 6}{RT_1}=3\times \dfrac{2P_1V_2}{RT_1}$

6 = 3 x 2 V₂

V₂= 1 L

So the final volume will be 1 L

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A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

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Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$

The value for static friction is $\mu_k = 0.70$

Explanation:

From the question we are told that

The mass of the clock is $m = 95 \ kg$

The first horizontal force is $F _s = 560 \ N$

The second horizontal force is $F _k = 650 \ N$

Generally the static frictional force is equal to the first horizontal force

So

$F _s = m * g * \mu_s$

=> $560 = 95 * 9.8 * \mu_s$

=> $\mu_s = 0.60$

Generally the kinetic frictional force is equal to the second horizontal force

So

$F _k = m * g * \mu_k$

$650 = 95 * 9.8 * \mu_k$

$\mu_k = 0.70$

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2 years ago

What are the factors that affect the size of an impact crater?

tatiyna

Answer:

The following factors could be responsible for size of an Impact Crater:

Geography of the region where it impacted.
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2 years ago

A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa

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Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

$mV_1+3mV_2=(m+3m)V_f=4mV_f$

$V_1+3V_2=4V_f$ and making $V_f$ the subject then

$V_f=\frac {V_1+3V_2}{4}$ and since $V_1$ is initial velocity of car, value given as 4 m/s, $V_2$ is the initial velocity of the three cars stuck together, value given as 2 m/s and $v_f$ is the final velocity which is unknown. By substitution

$V_f=\frac {4+(3\times2)}{4}=2.5 m/s$

(b)

Initial kinetic energy is given by

$\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ$

Final kinetic energy is given by

$\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ$

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

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3 years ago

Read 2 more answers

How do you find the water pressure at the bottom of the 55-m-high water tower?

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-- What's the volume of a cylinder with radius=1m and height=55m ?

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-- How much does that volume of water weigh ?

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-- What's the area of the bottom of that 1m-radius cylinder ?

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3 years ago

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Answer:

The answer is

<h2>10 m/s²</h2>

Explanation:

To find the acceleration of an object given the force and mass we use the formula

<h3> $acceleration = \frac{force}{mass}$ </h3>

From the question

mass of object = 50 kg

force = 500 N

So the acceleration is

<h3> $acceleration = \frac{500}{50} \\ = \frac{50}{5}$ </h3>

We have the final answer as

<h3>10 m/s²</h3>

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3 years ago