Rutherford's model of the atom (ESAAQ) Rutherford carried out some experiments which led to a change in ideas around the atom. His new model described the atom as a tiny, dense, positively charged core called a nucleus surrounded by lighter, negatively charged electrons.
Answer: D)supersaturated
Explanation: Solubility is defined as the amount of solute in grams which can dissolve in 100 g of the liquid to form a saturated solution at that particular temperature.
At
, the solubility of
is 153g/100 ml.
Thus if 180 grams is dissolved, it contains more amount of solute than it can hold at that that temperature, and thus is supersaturated solution.
A saturated solution is a solution containing the maximum concentration of a solute dissolved in the solvent. The additional solute does not dissolve in a saturated solution.
An unsaturated solution is solution in which the solute concentration is lower than its equilibrium solubility.
A supersaturated solution is one that has more solute than it can hold at a certain temperature.
Answer: Yos
Explanation: Becouse i experimented that btw
I'd say diffraction since sound waves can bend around objects like corners. Let's say you're in the hallway and you can hear sound coming from a door. The sound waves diffract around the door and spread out into the hallway, making it possible for you to hear.
Also, you can hear it before you see it because light waves are shorter than sound waves and hardly diffract around doors.
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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