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)
The functions of angles are used to find unknown lengths or angles that can't be measured, in terms of known quantities. The trig functions of angles are ratios of lengths, so they're bare naked numbers without units.
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.
-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.
-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.
-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.
Answer:
1/i + 1/o = 1/f thin lens equation
i = 33 * 8.9 / (33 - 8.9) = 12.2 cm to right of first lens
27 - 12.2 = 14.8 cm to left of second lens
i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens
