A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is
1 answer:
The sphere’s Electric potential energy is 1.6*J
Given,
q=6. 5 µc, V=240 v,
We know that sphere’s Electric potential energy(E) = qV=6.5*=1.6*
J
The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.
In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."
A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?
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Answer:
block K = 29.39 J and spring #1 Ke = 360 J
Explanation:
In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work
= Ef - E₀
Let's look for the energies
Initial
E₀ = Ke = ½ k₁ x₁²
Final, this is just before starting to compress the spring
Ef = Ke = ½ m v²
The work of the rubbing force is
= -fr x
Let's write Newton's second law the y axis
N-W = 0
N = W
fr = μ N
fr = μ mg
Let's replace
-μ mg x = ½ m v² - ½ k₁ x₁²
v² = 2/m (½ k₁ x1₁² -μ mg x)
v² = 2/6 (½ 2000 0.6²2 - 0.5 6 9.8 1) = 1/3 (360 - 29.4)
v = 3.13 m / s
With this value we calculate the energy of the block
K = ½ m v²
K = ½ 6 3.13²
K = 29.39 J
Calculate eenrgy of the spring ke 1
Ke = ½ k₁ x₁²
Ke = ½ 2000 0.60²
Ke = 360 J
The smallest perimeter of the rectangle is of value 150 cm.
Given:
The area of the rectangle, A = 1350 cm²
Calculation:
Let the length of the rectangle be 'x'
the breadth of the rectangle be 'y'
We know that the area of a rectangle is given as:
A = (x) × (y)
Applying values in the above equation we get:
xy = 1350 cm²
Factorizing the value of 1350, the possible values of length and breadth of the rectangle is as listed below:
x (cm) y (cm)
1350 × 1
675 × 2
450 × 3
270 × 5
225 × 6
150 × 9
135 × 10
90 × 15
75 × 18
54 × 25
50 × 27
45 × 30 (least possible value)
Thus, the smallest perimeter of the rectangle can be calculated as:
P = 2 (x + y)
= 2 (45 + 30)
= 150 cm
Therefore, the smallest perimeter that the rectangle will have is 150 cm.
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