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:
Answer:
A
Explanation:
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The correct option is A.
Answer:
Human-driven changes in arrive utilize and arrive cover such as deforestation, urbanization, and shifts in vegetation designs moreover change the climate, coming about in changes to the reflectivity of the Soil surface (albedo), emanations from burning timberlands, urban warm island impacts and changes within the normal water cycle.
The acceleration of the box up the ramp is 9.65 m/s².
<h3>
What is the magnitude of acceleration of the box?</h3>
The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;
F(net) = ma
where;
- m is the mass of the box
- a is the acceleration of the box
The net force on the box is calculated as follows;
F(net) = F - Ff
F(net) = F - μmgcosθ
where;
- θ is the inclination of the plane
- μ is coefficient of friction
F(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.7
The acceleration of the box is calculated as;
a = F(net) / m
a = (144.7) / (15)
a = 9.65 m/s²
Thus, the acceleration of the box up the ramp is 9.65 m/s².
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Answer:
See below
Explanation:
At point A the PE = mgh = 2 * 10 * 1 = 20 J
at point B, all of the PE , 20 J , is converted to Kinetic Energy
KE = 1/2 m v^2
20 = 1/2 (2)(v^2 )
20 = v^2 v = sqrt 20 = 4.47 m/s
for the friction part
vf = vo t + 1/2 a t^2 vf = final velocity = 0 (stopped)
vo = original velocity = 4.47 m/s
a = -1 m/s^2
0 = 4.47 t + 1/2 (-1) t^2
- .5t^2 + 4.47 t = 0
t ( -.5t+ 4.47) = 0 shows t = 4.47/.5 = 8.9 seconds
