Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the <span>friction force keeping on table is given by the following formula:</span>
<span>Ff = u*(1-f)*m*g </span>
and we need to know aso that <span>gravity force pulling off the table Fg is given by this other formula:</span>
<span>Fg = f*m*g </span>
What you need to do is <span>Equate the two and solve for f: </span>
<span>f*m*g = u*(1-f)*m*g </span>
<span>=> f = u*(1-f) = u - uf </span>
<span>=> f + uf = u </span>
=> f = u/(1+u) = fraction of rope
With that you can find the answer
Iodine is the answer to your question buddy
A.) kiloliter. 1 kiloliter = 1,000 liters
c.) megaliter. 1 megaliter = 1,000,000 liters
hope this helps
Answer:
0.000109375 m
Explanation:
d = Distance between grating = 0.5 mm
m = Order
Minima relation
For fourth order minima
For second maxima
From the two equations we get
The wavelength is 0.000109375 m
