A sealed cubical container 10.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Fin
1 answer:
This question involves the concepts of general gas equation and pressure.
The force exerted by the gas on one of the walls of the container is "74.08 KN".
First, we will use the general gas equation to find out the pressure of the gas:
where,
P = Pressure of the gas = ?
V = Volume of cube = (side length)³ = (10 cm)³ = (0.1 m)³ = 0.001 m³
n = no. of moles = 3 (since molecules equal to avogadro's number make up 1 mole)
R = general gas constant = 8.314 J/mol.K
T = Absolute Temperature = 24°C + 273 = 297 K
Therefore,
P = 7407.78 KPa
Now, the force on one wall can be given as follows:
where,
A = area of one wall = (side length)² = (0.1 m)² = 0.01 m²
Therefore,
<u>F = 74.08 KN</u>
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Learn more about the general gas equation here:
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Answer:
V = 0.714m/s
Explanation:
Full solution calculation can be found in the attachment below.
From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.
Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.
See the attachment below for the solution calculation.
<u>Answers:</u>
3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:
a) Here the ball is under 2 forces:
F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests
-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).
In this case the sum of the forces is 0
b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:
F2, the force of the strike
c) While the ball is in its flight, it is under the following forces:
F1, the force of the lift through the air
-F2, the gravity force
-F3, the force of air resistance, also called drag
F4, the tangencial force of the ball flight
4. Here are the sizes and directions of the resultant forces:
i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:
F=-10N+10N=0 The resulting force is zero
ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:
F=30N-40N=-10N This means the resulting force is 10N applied downwards
iii) In this case the only force applied to the block is -5N applied downwards
iv) Here there are four forces applied to the block.
In the y-axis we have to forces of the same size but opposite directions:
F1=10N-10N=0 This means the applied force in the y-axis is zero
In the x-axis we have two forces of different size and opposite directions:
F2=-15N+10N=-5N This means the resulting force is applied to the –x-side
