The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.
If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is
W = k[{x₂-x₁)² - (x₁-xn)²],
where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁
In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.
Then, substituting the values,
2 J = k (0.11² -0²)
k = 165.29 N/m
Finally, we use the value of k to the Hooke's Law to determine the Force.
F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
Answer:
Check the explanation
Explanation:
(a) When the electric field is applied to the ends of the wire, electrons move through the wire because the resistance of the wire is less than the resistance of the surrounding air. Resistance is the obstruction to the flow of electrons hence electrons flow through the path having minimum resistance.
(b) As the thermal energy increases, Kinetic energy of electrons as well as that of positive ions also increase. Hence the mean time between collisions of electrons with ions is reduced due to movement of ions also. That is why Resistance of the wire increases.
Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = 
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m 
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = 
calculate
f = 
f = 5.3 Hz
It us 200 energy if it the right awnser
