Answer:
F= σ² L² /2ε₀
F = (L² ε₀/4π) ΔV² / r⁴
Explanation:
a) For this exercise we can use Coulomb's law
F = - k Q² / r²
where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates
Capacitance is defined by
C = Q / ΔV
Q = C ΔV
also the capacitance for a parallel plate capacitor is related to its shape
C = ε₀ A / r
we substitute
Q = ε₀ A ΔV / r
we substitute in the force equation
F = k (ε₀ A ΔV / r)² / r²
k = 1 / 4πε₀
F = ε₀ /4π L² ΔV² / r⁴4
F = L² ΔV² ε₀/ (4π r⁴)
F = (L² ε₀/4π) ΔV² / r⁴
b) Another way to solve the exercise is to use the relationship between the force and the electric field
F = q E
where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface
Ф = ∫E .dA = 
/ ε₀
the plate have two side
2E A = q_{int} / ε₀
E = σ / 2ε₀
σ = q_{int} / A
substituting in force
F = q σ / 2ε₀
the charge total on the other plate is
q = σ A
q = σ L²
F= σ² L² /2ε₀
To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are
Comet mass 
Radius 
Rock was dropped from a height 'h' from surface = 1m
The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is
Where G means gravitational universal constant and M the mass of the planet
Now calculate the value of the time
The time taken for the rock to reach the surface is t = 87.58s
Answer:
The answer is: c. It does not move
Explanation:
Because the gravitational force is characterized by being an internal force within the Earth-particle system, in this case, the object of mass M. And since in this system there is no external force in the system, it can be concluded that the center of mass of the system will not move.
Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
+ 
+ 
= 
+
Ф
60 ·8 + 0 = (60 + 80)
Ф
480 = 140
Ф................. (I)
y-axis component form (+y north);
+ 
+ 
= 
+ 
Ф
0 + 80.9 = (60 + 80)
Ф
720=
140Ф
140Vf=
Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.
Answer:
Explanation:
b ) First is concave lens with focal length f₁ = - 12 cm .
object distance u = - 20 cm .
Lens formula
1 / v - 1 / u = 1 / f
1 / v + 1 / 20 = -1 / 12
1 / v = - 1 / 20 -1 / 12
= - .05 - .08333
= - .13333
v = - 1 / .13333
= - 7.5 cm
first image is formed before the first lens on the side of object.
This will become object for second lens
distance from second lens = 7.5 + 9 = 16.5 cm
c )
For second lens
object distance u = - 16.5 cm
focal length f₂ = + 12 cm ( lens is convex )
image distance = v
lens formula ,
1 / v - 1 / u = 1 / f₂
1 / v + 1 / 16.5 = 1 / 12
1 / v = 1 / 12 - 1 / 16.5
= .08333- .0606
= .02273
v = 1 / .02273
= 44 cm ( approx )
It will be formed on the other side of convex lens
distance from first lens
= 44 + 9 = 53 cm .
magnification by first lens = v / u
= -7.5 / -20 = .375 .
magnification by second lens = v / u
= 44 / - 16.5
= - 2.67
d )
total magnification
= .375 x - 2.67
= - 1.00125
height of final image
= 2.50 mm x 1.00125
= 2.503mm
e )
The final image will be inverted with respect to object because total magnification is negative .
