Answer: Improvement Invention means any CCIA Invention and CCIA's rights as a joint owner in a Joint Invention that is sufficiently different from the scope of a Licensed Patent to be separately patentable, and covered by the claims of Licensed Patents.
Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
P = S.G*p_w*g*h
Where, h = P / S.G*p_w*g
- Input the values given:
h = 101.325 KPa / 0.739*9.81*997
h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.
Icy/Snowy roads have less friction than normal roads. This means that the wheels are less likely to stay positioned because of traction, and you will spin out of control
Answer:
0.015 m/s2
Explanation:
Using Newtons 2nd law.
F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.
So substitute the values in SI units.
m = 
kg
Therefore F = 0.003×5 = 0.015 m/s2
Answer:
a) Temperatura, b) Temperature, c) Constant
, d) None of these
, e) Gibbs enthalpy and free energy (G)
Explanation:
a) the expression for ideal gases is PV = nRT
Temperature
b) The internal energy is E = K T
Temperature
c) S = ΔQ/T
In an isolated system ΔQ is zero, entropy is constant
Constant
d) all parameters change when changing status
None of these
e) Gibbs enthalpy and free energy
