20 POINTS!!!
1 answer:
a) Average velocity for her run north: 2.68 m/s north
b) The average velocity for the whole trip is zero
c) The average speed for the whole trip is 1.95 m/s
Explanation:
a)
In this part of the problem we only want to consider the part when Micha is running north.
The average velocity is given by:
where
d is the displacement (a vector connecting the initial position to the final position of motion)
t is the time
For the part running north, we have
north is the displacement
is the time interval
Substituting,
north (we have to specify also the direction, since velocity is a vector)
b)
As we said, average velocity is given by
where
d is the displacement
t is the time
If we consider the whole trip, however, the displacement is zero:
d = 0
Because Micha returns home, so the initial position corresponds to the final position of motion. Therefore, the average velocity is also zero:
v = 0
c)
The average speed is given by
where
d is the distance covered (the total length of the path, regardless of the direction)
t is the time interval
Since MIcha ran 2 miles north and then 2 miles back, the total distance is
And the time taken is
So, the average speed is
And since speed is a scalar, there is no need to specify a direction.
Learn more about speed and velocity:
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Answer:
The weight on the box and the normal force on the box will not be exactly opposite to one another. The angle between these two forces will be .
If the box isn't moving:
.
Explanation:
The weight on an object on the surface of the earth should always point downwards towards the center of the planet.
On the other hand, as the name suggest, the normal force on an object is normal to the surface that exerted this normal force. In this question, that surface is a plane with a incline to the ground. The corresponding normal force will be at
above the ground.
The angle between the normal force ( above the ground) and the weight (vertically downwards toward the ground.) would be
.
How could the box possibly not move even though the normal force on it isn't exactly opposite to its weight? Refer to the second diagram (not to scale) attached; decompose the weight on the box into two components:
- The first component is normal to the incline.
- The second component is parallel to the incline.
Notice how the original weight and the two decomposed forces form a right triangle.
Component normal to the incline: .
Component parallel to the incline: .
Notice that the vector sum of these two components is equal to the downward weight before the decomposition.
The normal force would be opposite to the component of the weight that is normal to the incline.The two forces should be equal in size. Therefore, the magnitude of the normal force should also be .
If the box is at equilibrium, then the friction on this box (parallel to the incline) should be equal in size to the component of weight that is parallel to the incline.
