Answer:
Explanation:
You didn't last any of the variables. You have to list the variables to tell which are which.
Answer:
No, its not possible for water to dissolve almost anything in the universe.
Explanation:
Solubility of a solute defines the ability of that solute to dissolve in a given solvent. It is defined as the maximum amount of solute dissolved in a solvent at equilibrium. The solution which results from dissolving this maximum amount is called a saturated solution, and one it has been reached, no more solute can be dissolved in it.
Different substances in the universe have diffferent solubilities in water, some very high (soluble) (eg. sugar and salt) and some very low (insoluble) (eg plastics). The substances that are able to form bonds with water (Hydrogen or Ionic) are more soluble than those who are not able to do so.
Answer:
Please see below as the answers are self-explanatory
Explanation:
a)
- A electric field line is an imaginary line, which has the property that the electric field vector is tangent to it at any point. It starts from positive charges (since the electric field by convention it has the direction of the trajectory that would take a positive test charge, so it always goes away from positive charges) and ends in negative charges.
b)
- Since the potential difference between two points represents the work per unit charge needed for a charge to move between these points, a equipotential surface is the one over which it is not needed to do work to move a charge from any point on the surface to any other point, which means that all points are at the same potential.
c)
- Equipotential surfaces are not necessarily physical surfaces, they can be defined in vaccum for instance.
- As an example, any spherical surface concentric with a point charge, is an equipotential surface, and it can be a real surface or a fictitious one.
Answer:
See Explanation
Explanation:
The relationship between angle of an incline and the acceleration of an object moving down the incline.
As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object
Parallel force = mgsin, perpendicular = mgcosΘ
With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.
mgsinΘ = ma
Fnet = ma
B.) From Fnet = ma
Fnet = ma
a = Fnet / m
Where Fnet = Net force = mgsinΘ, a = acceleration
