First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,

. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:

Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:

Now we can use the following relationship to find the distance covered by the skier before stopping, S:

where

is the final speed of the skier and

is the initial speed. Substituting numbers, we find:
Because they behave just like all the electromagnetic waves of the spectrum. Same equations, just shorter wavelengths and more energy.
Hope you get it :)
Answer:
9.73 x 10⁻¹⁰ m
Explanation:
According to Heisenberg uncertainty principle
Uncertainty in position x uncertainty in momentum ≥ h / 4π
Δ X x Δp ≥ h / 4π
Δp = mΔV
ΔV = Uncertainty in velocity
= 2 x 10⁻⁶ x 3 / 100
= 6 x 10⁻⁸
mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg
m = 9 x 10⁻¹⁹
Δp = mΔV
= 9 x 10⁻¹⁹ x 6 x 10⁻⁸
= 54 x 10⁻²⁷
Δ X x Δp ≥ h / 4π
Δ X x 54 x 10⁻²⁷ ≥ h / 4π
Δ X = h / 4π x 1 / 54 x 10⁻²⁷
= 
= 9.73 x 10⁻¹⁰ m
Answer:
105.8 m
46 m/s
Explanation:
From the time the rocket is launched to the time it reaches its maximum height:
v = 0 m/s
a = -10 m/s²
t = 9.2 s / 2 = 4.6 s
Find: Δy and v₀
Δy = vt − ½ at²
Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²
Δy = 105.8 m
v = at + v₀
0 m/s = (-10 m/s²) (4.6 s) + v₀
v₀ = 46 m/s
Static electricity<span> is caused by the build up of </span>electrical<span> charges on the surface of objects, while </span>current electricity<span> is a phenomenon from the flow of electrons along a conductor. 2. When objects are rubbed, a loss and/or gain of electrons occurs, which results in the phenomenon of </span>static electricity<span>.</span>