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:
The railroad tracks will expand because the heat waves make them bigger
Answer:
Derivation of Conservation of Momentum
Applying Newton's third law, these two impulsive forces are equal and opposite i.e. is equal to the change in momentum of the first object. is equal to the change in momentum of the second object. This relation suggests that momentum is conserved during the collision.
Explanation:
Hope it helps!!!
Answer:
Chief Hopper
Explanation:
Mike travels at a constant speed of 3.1 m/s. To find how long it takes him to reach the school, we need to find the distance he travels. We can do this using Pythagorean theorem.
a² + b² = c²
(1000 m)² + (900 m)² = c²
c ≈ 1345 m
So the time is:
v = d / t
3.1 m/s = 1345 m / t
t ≈ 434 s
Next, Chief Hopper travels a total distance of 1900 m, starting at rest and accelerating at 0.028 m/s². So we can use constant acceleration equation to find the time.
d = v₀ t + ½ at²
1900 m = (0 m/s) t + ½ (0.028 m/s²) t²
t ≈ 368 s
So Chief Hopper reaches the school first, approximately 66 seconds before Mike does.
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
Box with mass 2M
Box with mass 3M
On the third equation, acceleration can be modelled in terms of F'':
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
Afterwards, F' as function of the external force can be obtained by direct substitution:
The net forces of each block are now calculated:
Box with mass M
Box with mass 2M
Box with mass 3M
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.
