The coefficient of friction between the soap and the floor is 0.081
If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.
We know that frictional force F = μN where μ = coefficient of friction between soap and floor.
So, μ = F/N
Since F = 40 N and N = W = 493 N,
μ = F/N
μ = 40 N/493 N
μ = 0.081
So, the coefficient of friction between the soap and the floor is 0.081
Learn more about coefficient of friction here:
brainly.com/question/13923375
Answer:
The angular speed after 6s is
.
Explanation:
The equation

relates the moment of inertia
of a rigid body, and its angular acceleration
, with the force applied
at a distance
from the axis of rotation.
In our case, the force applied is
, at a distance
, to a ring with the moment of inertia of
; therefore, the angular acceleration is



Therefore, the angular speed
which is

after 6 seconds is


Hope this helps!!!!!!!!!!!!!
Answer:
114.86%
Explanation:
In both cases, there is a vertical force equal to the sprinter's weight:
Fy = mg
When running in a circle, there is an additional centripetal force:
Fx = mv²/r
The net force is found with Pythagorean theorem:
F² = Fx² + Fy²
F² = (mv²/r)² + (mg)²
F² = m² ((v²/r)² + g²)
F = m √((v²/r)² + g²)
Compared to just the vertical force:
F / Fy
m √((v²/r)² + g²) / mg
√((v²/r)² + g²) / g
Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:
√((12²/26)² + 9.8²) / 9.8
1.1486
The force is about 114.86% greater (round as needed).
When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling
the capacitance of the capacitor in air, the charge Q, the capacitance
and the voltage (
) are related by
(1)
when the source is disconnected the charge Q remains on the capacitor.
When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

And since
, substituting into the previous equation, we find:
