Let q(t) represent the amount of a certain reactant present at time t. suppose that the rate of decrease of q(t) is proportional
1 answer:
The rate of change is mathematically described as :
using variable seperation bring d(q) and all functions of q to one side and dt and all functions of t to the other side. It should look like this:
Now integrate on both sides.
where C is constant of integration. Its value can be found using initial condition by simply putting t=0 in the above equation
C =
After some algebraic simplification q(t) can be written as :
at half time, q(T) = q(0)/2, where T = half time of reaction.
substitute this in the above equation:
This is the required solution. As can be seen half time (T) depends on K and q(0) that is the initial concentration.
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here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.
So angle must be more than 270 degree and less than 360 degree
Now in order to find the value we can say that
so it is inclined at above angle with X axis in fourth quadrant
Now if angle is to be measured counterclockwise then its magnitude will be
so the correct answer will be 305 degree
Answer:
(a) 0.0113 ±0.0001 inches
(b) 5 decks
Explanation:
<u>Given information</u>
52 cards, thickness is 0.590 ±0.005 inches
Since this is thickness of all the cards, to get the thickness of a single card we divide the total thickness (plus uncertainty) by the number of cards hence
1 card=±
= 0.0113461 ±0.000096 inches
Considering that thickness is given to 3 significant figures while uncertainty is to 1 significant figure, the final answer should also conform to these hence giving the first part of the answer to 3 significant figures while second part to 1 significant figure yields 0.0113 ±0.0001 inches
(b)
Considering that the cards have uncertainty of 0.0001 inches and the number of decks, n required to create uncertainty of 0.00002 inch is given by
We need 5 decks for the given uncertainty
Answer:
a.
b. m=5.99kg
Explanation:
a.
In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).
In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.
So we start with the sum of forces about y, so we get:
N-W=0
N=W
N=mg
Next we can do the sum of forces about x, so we get:
which yields:
so:
Next the torque about point B, so we get:
so:
From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:
We can now combine like terms, so we get:
we know that W=mg, so we can substitute that into our equation, so we get:
which can now be solved for the mass m:
If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:
we can now divide both sides of the equation into g so we get:
next we can divide both sides of the equation into cos θ so we get:
and finally we can multiply both sides of the equation by 2 so we get:
we know that:
so we can simplify the equation a little more, so we get:
b.
So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:
θ=32° and
we also know that
so we can use our equation now:
so we get:
which yields:
m=5.99kg
