Wait I take that Batrachotoxinin A does not have carboxylic acid any more, but a hydroxy group, so it's not an ester, but an alcohol.
So I somehow have to name it with a substituted ethanol??
Any hints??
So I came across a cool toxin, and saw that people have not assigned an IUPAC name to it, so I took on the challenge.
The compound of interest is Batrachotoxinin A
unc.
edu/depts/mtcgroup/litmeetings/batrachotoxinin.pdf
In the second slide Batrachotoxinin A has an R which is an H...
To those of you who don't see from the implication, I shall expand.
Look carefully at the circuit, the three emf on the left of the circuit are moving the charge in the clockwise direction, and the one emf on the right is the opposite direction.
So for the sum of emf, don't forget to make...
Sorry I didn't specify a battery. Let's just say a D cell alkaline battery with 1.5V. And by weak, I mean, nearly discharged, or nearly depleted in chemical charge due to the extensive oxidation reaction at the anion.
You see what I'm getting confused is that I ask someone about it and they say...
Relevant equations:
V = I*R
ε = I*(R+r)
Then,
V = {ε/(R+r)}*r
V = potential difference, ε = electromotive force, R = resistance of external load resistor,
r = internal resistance of the battery.
So say the external load resistor's resistance is a constant for two batteries of the...
To simplify and sum it up, the emf of a cell is the total potential difference the cell can produce around the circuit, including any potential wasted in driving the current through the cell itself.
V = I*R
ε = I *(R+r)
I = V/R
ε = V/R *(R+r)
Where V is the potential difference, ε is the...
Thank you very much Dacruick for clearing that up. So my thought of the normal force being equal to the x-component of the force applied was the correct assumption.
Now the comment on the y-component of the force applied, was beneficial. You said "If the y component of the applied force...
Dacruick you are right, I'm imagining the book to be static. Sorry, I forgot to mention that. The net force is zero, but my question is on the vector component of the forces(specifically the relationship between the x-component of force applied and the normal force), and the angle of the force...
Consider that the book and the force applied by the hand is all together one part, so no worries if there is any slippage between the hand and the book. So then, the book is pressed against the wall with some force. The free-body diagram drawn in my head has a force of friction( μ static*normal...
I think I solved it... duh.
in the distance formula
d = v(initial)*t + 1/2*a*t^2.v(initial) and time is already attained and the the a acceleration is -9.81m/s^2
Correct me if I'm wrong.
Imagine a ramp setup on top of a tall table. The height Δy is measured. To find the initial velocity at the instant the ball leaves the ramp, I set up the kinetic energy and potential energy equal to each other to find the initial velocity of the x component.
PE = KE
m*g*(hr) = 0.5*m*v^2
where...