distance of each pan from the center or fulcrum is given as

now if dishonest shopkeeper shifted it by 0.633 cm from center
so distance on each side is given as


now the weight is balance as


now we will have

now we can find the percentage change as


Answer:
a = 0.009 J
b = 0.19 m/s
c = 0.005 J and 0.004 J
Explanation:
Given that
Mass of the object, m = 0.5 kg
Spring constant of the spring, k = 20 N/m
Amplitude of the motion, A = 3 cm = 0.03 m
Displacement of the system, x = 2 cm = 0.02 m
a
Total energy of the system, E =
E = 1/2 * k * A²
E = 1/2 * 20 * 0.03²
E = 10 * 0.0009
E = 0.009 J
b
E = 1/2 * k * A² = 1/2 * m * v(max)²
1/2 * m * v(max)² = 0.009
1/2 * 0.5 * v(max)² = 0.009
v(max)² = 0.009 * 2/0.5
v(max)² = 0.018 / 0.5
v(max)² = 0.036
v(max) = √0.036
v(max) = 0.19 m/s
c
V = ±√[(k/m) * (A² - x²)]
V = ±√[(20/0.5) * (0.03² - 0.02²)]
V = ±√(40 * 0.0005)
V = ±√0.02
V = ±0.141 m/s
Kinetic Energy, K = 1/2 * m * v²
K = 1/2 * 0.5 * 0.141²
K = 1/4 * 0.02
K = 0.005 J
Potential Energy, P = 1/2 * k * x²
P = 1/2 * 20 * 0.02²
P = 10 * 0.0004
P = 0.004 J
Ok so the first thing you need to do is im not gonna help you
Explanation:
What characteristics must the movement of a person have so that the value of the displacement is equal to the distance traveled?
Displacement is equal to the shortest path covered by an object. It is given by the difference of final position and the initial position.
Distance is equal to the total path covered by an object during the journey.
When an object moves in a straight line path, in this case, the displacement is equal to the distance traveled.
Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal