Answer:
a) 
b)
º
c) 
Explanation:
From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together
So, If we analyze the collision we got

To simplify the problem, lets name D for Daniel and R for Rebecca
a) 
Since Daniel's initial velocity is 0



Now, lets analyze the movement in the vertical direction

Since 


Now, we can find the magnitude of Daniel's velocity after de collision

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle
º
c) The change in the total kinetic energy is:
ΔK=
ΔK=![\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%2845kg%29%288m%2Fs%29%5E2%2B%2870kg%29%287.32m%2Fs%29%5E2-%2845kg%29%2814m%2Fs%29%5E2%5D%3D-1094.62J)
That means that the kinetic energy decreases
The change in the Gibb's free energy per mole (G) is 1.96 J.
The given parameters:
- Density of the ice, ρ = 917 kg/m³
- Initial pressure, P₁ = 1.0 bar
- Final pressure, P₂ = 2.0 bar
- Temperature, T = - 10 C
- Mass of water = 18 g
The change in the Gibb's free energy per mole (G) is calculated as follows;

where;
V is the volume of the ice

Change in pressure;

The change in the Gibb's free energy per mole (G);

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.
Learn more about Gibb's free energy here: brainly.com/question/10012881
Answer:
The time taken is 
Explanation:
From the question are told that
The speed of the incoming water is 
The cross-sectional area of the host pipe is
Generally the rate at which the container is been filled is mathematically represented as

=> 
=> 
Generally 
Generally the time taken is mathematically represented as

=> 
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g