When you fall, you arrive at the floor with some momentum. It makes
no difference <em>what</em> stops your fall, or how long it takes to stop your fall.
It still has to absorb the same amount of momentum, somehow. As long
as your body still has any downward momentum, you're still falling.
The total change in momentum can also be viewed as some amount
of 'impulse' ... the product of force and the time that the force persists.
The knee pads absorb your momentum slower than the bare floor does,
and since the process of changing your momentum lasts longer, the
corresponding <em>force</em> against your knees is smaller with the pads.
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s
Answer:
The voltage across the capacitor decreases by a factor of 2
Explanation:
When we put a dielectric material on a capacitor, we generate a polarization on it that changes the capacitance of the capacitor. The dielectric constant is the ratio between the initial capacitance (Co) without the dielectric and the new capacitance (C) with the dielectric:
![k=\frac{C}{C_0}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BC%7D%7BC_0%7D%20)
So, if k is equal to 2:
![C=2*C_0](https://tex.z-dn.net/?f=C%3D2%2AC_0%20)
So, the capacitance is doubled with the dielectric
Because the battery is disconnected, the charge on the plates of capacitor must be constant because conservation of charge.
The effective electric field of a capacitor is the electric filed when we put a dielectric on it, and it is:
![E_{eff}=\frac{E}{k}=\frac{E}{2}](https://tex.z-dn.net/?f=E_%7Beff%7D%3D%5Cfrac%7BE%7D%7Bk%7D%3D%5Cfrac%7BE%7D%7B2%7D%20)
With E the initial electric field, so the electric field is halved
Finally, because we know electric field and voltage (V) on a parallel capacitor with distance between plates (d) are related by:
![E=\frac{V}{d}](https://tex.z-dn.net/?f=%20E%3D%5Cfrac%7BV%7D%7Bd%7D)
Because electric field and voltage are directly proportional and d remains constant if Electric field is halved, then voltage is halved too. The voltage across the capacitor decreases by a factor of 2.