Answer:
JRJJEJERJRJERERJREREJERJJERJERTJE
ExpJERlanation:
SDSHERHJRESHERDHEDGERJEJERJERJERRJERSH
Answer:
488.6KN
Explanation:
Hello!
the first step to solve this problem we must find the pressure exerted at the bottom of the tank (P) which is the sum of the external air pressure (P1 = 92kPa), the pressure inside the tank (P2 = 100kPa) and the pressure due to the weight of the water (P3), taking into account the above we have the following equation
P=P1+P2+P3
to find the pressure at the bottom of the tank due to the weight of the water we use the following equation

where
α=density=1 g/cm^3=1000kg/M^3
H=height=14.1m
g=gravity=3.71m/s^2
solving
P3=(1000)(14.1)(3.71)=52311Pa=52.3kPa
P=P1+P2+P3
P=100kPa+92kPa+52.3kPa=244.3kPa
finally to solve the problem we remember that the pressure is the force exerted on the area

Answer:
p = 20 kg•m/s
KE = 100 J
Explanation:
In an elastic collision of identical masses, the two masses will exchange momentums. Therefore Block 1 initially moving at 10 m/s will be moving at 2 m/s, and Block 2 will go from 2 m/s to 10 m/s
momentum = mv = 2(10) = 20 kg•m/s
KE = ½mv² = ½(2)10² = 100 J
Unfortunately, your answer selection does not have this answer as an option.
Answer:
The correct option is;
a. Any process in which the entropy of the universe increases will be product-favored
Explanation:
According to the second law of thermodynamics, the change in entropy of a closed system with time is always positive. That is the entropy of the entire universe, considered as an isolated system, always increases with time, hence the entropy change in the universe will always be positive.

Therefore, any process in which the entropy of the universe increases will be product favored.
Is potential energy that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system