Answer:
As the roller coaster goes higher, more potential energy is increased in the roller coaster. Caused by gravity and the roller coaster's position, the potential energy is stored in the roller coaster. For example, this ball is at the top of a hill, where potential energy is at it's highest. Potential energy can be calculated by Potential Energy=Mass X Acceleration X Height.
Explanation:
Answer:
0.5 Ohms
Explanation:
We note that the node Q is also between the resistors of 1ohm and 2ohms.
We note that the node P is also between the resistors of 2ohms and 3ohms.
Thus, all these resistors are in parallel, beween nodes P and Q
1/Re=1/R1+1/R2+1/R3
1/Re=1/1+1/2+1/3=(6+3+2)/6=11/6 [ohm^(-1)]
Re=6/11=0.54ohms
Rounding to the tenth: Re=0.5 ohms
Answer:
The displacement is zero
Explanation:
Displacement is a vector that connects the starting position of motion of an object to the final position. It is a vector quantity, so it has both magnitude and direction.
Viceversa, distance is a scalar quantity that measures the total length of the path covered by the object during its motion.
In this problem, we are interested in the displacement.
We are told that the runner goes around a 1 mile track, and he ends at his starting position: this means that his final position is equal to the starting position, therefore the displacement is zero.
Answer:
First Dot That is located on Earth.
Explanation:
Due to Earth's gravity being greater than the moon's on earth there is more gravitational force pulling down on the spacecraft.
Answer:
snow
Explanation:
Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.
We can sole this problem using the following derivation:
![ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BV_f%7D%7BV_i%7D%20%29%3D-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%5Cfrac%7BP_i%7D%7BP_f%7D%20%20%29%5C%5C%3D-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%29-%28%5Cgamma%20-1%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%3D-%28%5Cfrac%7B%5Cgamma%20-1%7D%7B%5Cgamma%7D%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%3D-%28%5Cfrac%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%20-1%7D%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%7D%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%5C%5Cln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D%3D-%28%5Cfrac%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%20-1%7D%7B%5Cfrac%7BC_%7Bp%2Cm%7D%7D%7BC_%7Bp%2Cm%7D-R%7D%7D%29ln%28%5Cfrac%7BP_i%7D%7BP_f%7D%29%5C%5C%5C%5CSubstituting%5C%20values%3A%5C%5C%5C%5C)
![ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D-%28%5Cfrac%7B%5Cfrac%7B28.86%7D%7B28.86-8.314%7D%20-1%7D%7B%5Cfrac%7B28.86%7D%7B28.86-8.314%7D%7D%29ln%28%5Cfrac%7B0.802%5C%20atm%7D%7B0.602%5C%20atm%7D%29%3D-0.0826%5C%5C%5C%5Cln%28%5Cfrac%7BT_2%7D%7BT_1%7D%20%29%3D-0.0826%5C%5C%5C%5CTaking%5C%20exponential%5C%20of%5C%20both%20%5C%20sides%3A%5C%5C%5C%5C%5Cfrac%7BT_2%7D%7BT_1%7D%20%3De%5E%7B-0.0826%7D%5C%5C%5C%5CT_2%3D0.9207T_1%5C%5C%5C%5CT_2%3D0.9207%2A288%5C%5C%5C%5CT_2%3D265%5C%20K%5C%5C)
Since T2 = 265 K, we should expect a snow