<span>First lets determine the equation. Well at the top of the circle both the normal force and the weight are in the same direction. So we have Fnet=N+mg. Since this is a circular path the Fnet is also = to (mv^2)/r.
We convert the situation where the rock is no longer in contact with the bottom to terms relevant to the equation. So, what is a requirement for normal force? The object must be in contact with the surface, meaning it can't be in free fall. Realizing this means that the instant when the object does not touch the bucket is where the normal force = 0.
Now we have N+mg=(mv^2)/r where N=0 is the case we are interested in. This leaves 0+mg=(mv^2)/r
Solve for v:
v=(gr)^(1/2) or v=3.28m/s
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Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
Explanation:
It is given that the value of P is 1110 N. And, for pin connection we have only two connections which are 
and 
. Let T be the tension is rope.
So, 
and - 1110 = 0
= 0
And, 
= 0
T =
Also, 
= 0
1110(0.75 + 0.75 + 0.75) - T(0.5 + 0.1) = 0
2497.5 - 0.6T = 0
T = 4162.5 N
= 4.16 kN
Therefore, we can conclude that the tension in the rope if the frame is in equilibrium is 4.16 kN.
Answer:
Okey is my opinion C anyway
That's what happens whenever anything is dropped or falls.
-- a roller coaster leaves the top of a hill
-- a pencil rolls off the desk
-- the cat jumps off the top of the refrigerator
-- the little boy slides down the slide
