Answer:
Explanation:
As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as
so here we know that
so we have
A solid solute is added to a liquid solvent. The solution is stirred and heated. How does stirring and heating affect the soluti
on?
A) Stirring increases the rate of dissolving, but heating decreases the rate.
B) Both stirring and heating increase the rate of dissolving.
C) Both stirring and heating decrease the rate of dissolving.
D) Stirring decreases the rate of dissolving, but heating increases the rate.
A) Stirring increases the rate of dissolving, but heating decreases the rate.
B) Both stirring and heating increase the rate of dissolving.
C) Both stirring and heating decrease the rate of dissolving.
D) Stirring decreases the rate of dissolving, but heating increases the rate.
2 answers:
Answer:
B) Both stirring and heating increase the rate of dissolving.
Explanation:
When we dissolve a solute in a solvent, it is separated into its constituting ions.
For example, if we dissolve salt in water, The NaCl is disassociated into its ions and
because of the interaction of
and
ions present in water.
if we can increase the amount of salt interacting with water, we can increase the rate of dissolving.
The picture below will provide a better understanding about the area of reaction.
In the figure a, salt is accumulated at the bottom, in this case only the salt molecules at the upper surface will be able to react with water.
In the figure b, salt grains are moving around the container. This increases the number of molecules interacting with water. This leads to quicker dissolving.
We can make the salt grains move by either heating or by stirring. So both stirring and heating will increase the rate of dissolving
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Answer:
Hence the answer is E inside .
Explanation:
E inside
so if r1 will be the same then
E proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle,
Coefficient of kinetic friction,
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:
Now, the speed is given by the third eqn of motion with initial velocity being zero:
where
u = initial velocity = 0
Thus