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3 years ago

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A piston-cylinder device initially contains 0.3 kg of nitrogen gas at 160 kPa and 140oC. The nitrogen is now expanded isothermal

ly to a pressure of 100 kPa. Determine the boundary work done during this process.

1 answer:

irakobra [83]3 years ago

4 0

Answer:

<h2>Work done by the gas is given as</h2><h2> $W = 1.72 \times 10^4 J$ </h2>

Explanation:

As we know that the process is isothermal so here work done is given as

$W = nRTln(\frac{P_1}{P_2})$

here we know that

$n = \frac{w}{M}$

$n = \frac{300}{28} = 10.7 moles$

now we have

$W = 10.7 (8.31)(273 + 140) ln(\frac{160}{100})$

so we have

$W = 1.72 \times 10^4 J$

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Tomtit [17]

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$

Learn more about electric force:

brainly.com/question/8960054

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3 years ago

I'll mark brainliest

irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

= 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

= 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

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An object is moving to the west at a constant speed. three forces are exerted on the object. one force is 10 n directed due nort

SVETLANKA909090 [29]

If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
$F_{12} = \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N$
with direction at $45^{\circ}$ north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be $45^{\circ}$ south-east.

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ExtremeBDS [4]

Explanation:

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It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
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v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

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an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr

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The horizontal velocity was constant, so:

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