Answer:
The force applied 275 N in a direction parallel to the hill
Explanation:
Newton's second law is adequate to work this problem, in the annex we can see a free body diagram, where the weight (W) is vertical, the friction force (fr) is parallel to the surface and the normal (N ) is perpendicular to it. In general for these problems a reference system is taken that is parallel to the surface and the Y axis is perpendicular to it.
Let us decompose the weight into its two components, the angle T is taken from the axis and
Wx = W sin θ
Wy = W cos T
We write Newton's second law
∑ F = m a
X axis
The cyclist falls at a constant speed, which implies that the acceleration is zero
fr - W sin θ = 0
fr = mg sin θ
fr = 96 9.8 without 17
fr = 275 N
When the cyclist returns to climb the hill, he must apply the same force he has to overcome the friction force that always opposes the movement
. The force applied 275 N in a direction parallel to the hill
Answer:
La frecuencia será la misma en los dos medios, y en el vacio, no varia.
b) Vacío : n = 3,684 10 m 1,33 = 4 900 Å ,3161 10 m 55,1 n
,6 105 10 Hz
3,161 10 m
v 1,93 10 m/s 193548,38 km/s =
1,55
10 3 m/s
n
c
Vidrio v: =
,6 107 10 Hz
3,684 10 m
v 2,25 10 m/s 225 000km/s =
33,1
10 3 m/s
n
c
a
This is a problem that can be solved using free-fall motion analysis. Since the displacement (2.9m) is given, we can use the following equation to solve for the impact speed:
V^2 = 2gh
V = sqrt (2*9.8*2.9)
V = 7.54 m/s
This is observational learning because Ian observed that his peers waited in the cafeteria until the first bell rings. He decided to imitate them.