Answer:
we approach a maximum or minimum the values of the ordinate are closer and closer and when passing this point the values change their trend
Explanation:
The reason for this process occurs because as we approach a maximum or minimum the values of the ordinate are closer and closer and when passing this point the values change their trend if they were rising, they begin to fall and if they were falling they begin to rise. Therefore the maximum point is a point of inflection of the curve since its trend changes.
Another way of looking at this process is that mathematically the point where there is a maximum or a minimum corresponds to the point where the first derivative is equal to zero, this is the slope of the line is horizontal, so the points before after correspond to values with slope of different sign.
The answer is no, the static charge would still be there.
Answer:
A. Mrŵ² = ųMg
Ŵ = (ųg/r)^½
B.
Ŵ =[ (g /r)* tan á]^½
Explanation:
T.v.= centrepetal force = mrŵ²
Where m = mass of block,
r = radius
Ŵ = angular momentum
On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.
So there for
Mrŵ² = ųMg
Ŵ = (ųg/r)^½
g = Gravitational pull
ų = coefficient of friction.
B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible
So therefore
N *(sin á) = mrŵ² .....equ 1
Since the car does not slide the net vertical forces must be equal and opposite so therefore
N*(cos á) = mg.....equ 2
Where N is the reaction force of the car on the surface.
Equ 2 becomes N = mg/cos á
Substituting N into equation 1
mg*(sin á /cos á) =mrŵ²
Tan á = rŵ²/g
Ŵ =[ (g /r)* tan á]^½
Answer:
B) The same as the momentum change of the heavier fragment.
Explanation:
Since the initial momentum of the system is zero, we have
0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.
0 = p + p'
p = -p'
Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p
The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'
Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'
<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment. </u>
So, option B is the answer
