Answer:
5 percent = normal matter
68 percent = dark energy
27 percent = dark matter
Explanation:
The mass affects the kinetic energy because the more the mass the more energy is given to the object and the speed<span> affects by making it go faster and longer, so whenever speed goes up so does energy.</span>
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,

. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:

Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:

Now we can use the following relationship to find the distance covered by the skier before stopping, S:

where

is the final speed of the skier and

is the initial speed. Substituting numbers, we find:
Answer:
when they have the same slope
Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg