Answer: A) Forces of attraction and repulsion exist between gas particles at close range.
Explanation:
The <u>Ideal Gas equation</u> is:
Where:
is the pressure of the gas
is the volume of the gas
the number of moles of gas
is the gas constant
is the absolute temperature of the gas in Kelvin
According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them. In this sense, real gases can behave approximately to an ideal gas, under conditions of high temperature and low pressures.
However, at low temperatures or high pressures, real gases deviate significantly from ideal gas behavior. This is because at low temperatures molecules begin to move slower, allowing the repulsive and attractive forces among them to take effect. In fact, <u>the attraction forces are responsible for the condensation of the gas</u>. In addition, at high pressures the volume of molecules cannot be approximated to zero, hence the volume of these molecules is not negligible anymore.
Answer:
Chemical reactions occur slower at lower temperatures and faster at higher temperatures. When you put a glow stick in cold water, the chemical reaction slows down but will last for a longer period of time. When you put a glow stick in hot water, the reaction speeds up but will be over quicker.
Explanation:
(self-explanatory)
I hope this helps, have a nice day.
Answer:
b) 6
Explanation:
Given
v(t)=3t²+6t
X(0) = 2
X(1) = ?
Knowing that
v(t)=3t²+6t = dX/dt
⇒ ∫dX = ∫(3t²+6t)dt
⇒ X - X₀ = t³ + 3t²
⇒ X(t) = X₀ + t³ + 3t²
If X(0) = 2
⇒ X(0) = X₀ + (0)³ + 3(0)² = 2
⇒ X₀ = 2
then we have
X(t) = t³ + 3t² + 2
when
t = 1
X(1) = (1)³ + 3(1)² + 2
X(1) = 6
The wave property is called frequency
Answer:
h = 40.37 m
Explanation:
We will apply the law of conservation of energy to the skier in this case, as follows:
where,
m = mass of skier = 77 kg
g = acceleration due to gravity = 9.81 m/s²
vf = final speed = 30 m/s
vi = initial speed = 2 m/s
W_friction = Work done by friction and air resistance = 4000 J
Therefore,
![(77\ kg)(9.81\ m/s^2)h = \frac{1}{2}(77\ kg)[(30\ m/s)^2-(2\ m/s)^2] - 4000\ J\\\\h = \frac{34496\ J - 4000\ J}{755.37\ N}\\\\](https://tex.z-dn.net/?f=%2877%5C%20kg%29%289.81%5C%20m%2Fs%5E2%29h%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2877%5C%20kg%29%5B%2830%5C%20m%2Fs%29%5E2-%282%5C%20m%2Fs%29%5E2%5D%20-%204000%5C%20J%5C%5C%5C%5Ch%20%3D%20%5Cfrac%7B34496%5C%20J%20-%204000%5C%20J%7D%7B755.37%5C%20N%7D%5C%5C%5C%5C)
<u>h = 40.37 m</u>
