Answer: ![-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BBr%5E.%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Explanation:
Rate of a reaction is defined as the rate of change of concentration per unit time.
Thus for reaction:
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{d[Br^.]}{2dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D)
or ![Rate=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Thus ![-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Well, first of all, there's no such thing as "fully charged" for a capacitor.
A capacitor has a "maximum working voltage", because of mechanical
or chemical reasons, just like a car has a maximum safe speed. But
anywhere below that, cars and capacitors do their jobs just fine, without
any risk of failing.
So we have a capacitor that has some charge on it, and therefore some
voltage across it. From the list of choices above . . .
<span>-- Both plates have the same amount of charge.
Yes. And both plates have opposite TYPES of charge.
One plate is loaded with electrons and is negatively charged.
The other plate is missing electrons and is positively charged.
-- There is a potential difference between the plates.
Yes. That's the "voltage" mentioned earlier.
It's a measure of how badly the extra electrons want to jump
from the negative plate to the positive plate.
-- Electric potential energy is stored.
Yes. It's the energy that had to be put into the capacitor
to move electrons away from one plate and cram them
onto the other plate.
</span>
"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."
These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is
"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."
The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.
The distance covered by the acorn is 3.136 m.
<u>Explanation:</u>
The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.
Then using the second law of equation,
Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so
So the distance covered by the acorn is 3.136 m.
