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:
T= 3.34
Vi= 0
A= 9.81
D= ?
d=Vit+1/2at^2
d= 1/2(9.81)(3.34)2
d= 54.7 or 55 meters tall
Answer:
E = 13.2 kWh
, Cost = $ 10.8
Explanation:
We can look for the consumed energy from the expression of the power
P = W / t
The work is equal to the variation of the kinetic energy, for which
P = E / t
E = P t
look for the energy consumed in one day and multiply by the days of the month in the month
E = 110 4 30
E = 13200 W h
E = 13.2 kWh
the cost of this energy is
Cost = 0.9 12
Cost = $ 10.8
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 +
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
Explanation: (I think)
Plug your values into the momentum equation.
So m1= 63kg
m2 = 10 kg
V1 = 12 m/s
And then plug in your values and solve for your unknown (v2)