Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
Answer:
The angular speed of the object is 0.0281 rad/s
The linear speed of the object is 0.169 ft/s
Explanation:
Given;
radius of the circle, r = 6 ft
time of motion of the object around the circle, t = 80 s
central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad
The angular speed of the object is calculated as;
The linear speed of the object is calculated as;
v = ωr
v = 0.0281 rad/s x 6ft
v = 0.169 ft/s
The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
Answer:
Explanation:
We have to find electric potential V at a distance r.
a) For r>R,
The electric field in the cylinder is given by
E.A equating it to the other electric field given by
б.A/ε₀
Here the area of cylinder is given by= 2*3.14*r*L
While for the outside, the area= 2*3.14*R*L
Equating both, we get
E= бR/rε₀
Now,
The potential difference is given as:
ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.
Where ΔV= V₀-V
So solving we get
V₀=V-бR/ε₀ln (r/R)
b) For r<R i.e. inside the cylinder
There will be no electric field produced as E=0
So ultimately Vin= V
c) V=0 at r= infinity.
There must be movement in the same direction as the force put on the object. Hope this helps!
