The Ideal Gas Law makes a few assumptions from the Kinetic-Molecular Theory. These assumptions make our work much easier but aren't true under all conditions. The assumptions are,
1) Particles of a gas have virtually no volume and are like single points.
2) Particles exhibit no attractions or repulsions between them.
3) Particles are in continuous, random motion.
4) Collisions between particles are elastic, meaning basically that when they collide, they don't lose any energy.
5) The average kinetic energy is the same for all gasses at a given temperature, regardless of the identity of the gas.
It's generally true that gasses are mostly empty space and their particles occupy very little volume. Gasses are usually far enough apart that they exhibit very little attractive or repulsive forces. When energetic, the gas particles are also in fairly continuous motion, and without other forces, the motion is basically random. Collisions absorb very little energy, and the average KE is pretty close.
Most of these assumptions are dependent on having gas particles very spread apart. When is that true? Think about the other gas laws to remember what properties are related to volume.
A gas with a low pressure and a high temperature will be spread out and therefore exhibit ideal properties.
So, in analyzing the four choices given, we look for low P and high T.
A is at absolute zero, which is pretty much impossible, and definitely does not describe a gas. We rule this out immediately.
B and D are at the same temperature (273 K, or 0 °C), but C is at 100 K, or -173 K. This is very cold, so we rule that out.
We move on to comparing the pressures of B and D. Remember, a low pressure means the particles are more spread out. B has P = 1 Pa, but D has 100 kPa. We need the same units to confirm. Based on our metric prefixes, we know that kPa is kilopascals, and is thus 1000 pascals. So, the pressure of D is five orders of magnitude greater! Thus, the answer is B.
As the metal expands as does the road bed so neither really effevts those foing over the bridge. as it is hot the metal will expand and so will most tarmac on roads.
Answer:
The coefficient of static friction between the puppy and the floor is 0.7273.
Explanation:
The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:
Where 
is the static friction force, 
is the coefficient of static friction and 
is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, 
, to make the puppy moving we need to use a force of 80 N, therefore, 
, so we can solve for the coefficient as shown below:
The coefficient of static friction between the puppy and the floor is 0.7273.
Answer:
a) Revolutions per minute = 2.33
b) Centripetal acceleration = 11649.44 m/s²
Explanation:
a) Angular velocity is the ratio of linear velocity and radius.
Here linear velocity = 72 m/s
Radius, r = 0.89 x 0. 5 = 0.445 m
Angular velocity
Frequency
Revolutions per minute = 2.33
b) Centripetal acceleration
Here linear velocity = 72 m/s
Radius, r = 0.445 m
Substituting
Centripetal acceleration = 11649.44m/s²
<span>(1) </span>Through the Second
Law of motion, the equation for Force is:
F = m x a
Where
m is mass and a is acceleration (deceleration)
<span>(2) </span>Distance is
calculated through the equation,
D
= Vi^2 / 2a
Where
Vi is initial velocity
<span>(3) </span>Work is calculated
through the equation,
W = F x D
Substituting
the known values,
Part
A:
<span>(1) </span> F = (85
kg)(2 m/s^2) = 170 N
<span>(2) </span> D = (37
m/s)^2 / (2)(2 m/s^2) = 9.25 m
<span>(3) </span> W = (170
N)(9.25 m) = 1572.5 J
Part
B:
<span>(1) </span> F = (85 kg)(4
m/s^2) = 340 N
<span>(2) </span>D = (37 m/s)^2 /
(2)(4 m/s^2) = 4.625 m
<span>(3) </span><span> W = (340
N)(4.625 m) = 1572.5 J</span>
