Answer:

Explanation:
The equation for kappa ( κ) is

we can find the maximum of kappa for a given value of b using derivation.
As b is fixed, we can use kappa as a function of a

Now, the conditions to find a maximum at
are:


Taking the first derivative:








This clearly will be zero when

as both are greater (or equal) than zero, this implies

The second derivative is




We dcan skip solving the equation noting that, if a=b, then

at this point, this give us only the first term

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum
the value of kappa is



Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.
<span>fast-moving particles colliding with slow-moving particles</span>