How can two tectonic plates move relative to each other
1 answer:
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Answer:
a. They will be tie
b. Win the wood cylinder
Explanation:
a.
The both cylinders will reach the bottom at the same time notice the relation in the equation in indepent of the length and both have the same radius and the same rotational inertia.
So both will be tie
b.
The acceleration of the wood cylinder is larger than the acceleration of the brass cylinder so the cylinder of wood will reach the bottom first
So the wood win the race
Answer:
The speed at which the ball rolled off the end of the table is 3.3 m/s
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.
The position vector of the ball can be calculated as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity.
When the ball reaches the ground, its position will be:
r final = (3, -4)
Then:
3 = x0 + v0x · t
-4 = y0 + v0y · t + 1/2 · g · t²
Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:
3 m = v0x · t
-4 m = 1/2 · g · t²
We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:
-4 m = 1/2 · g · t²
-4 m = -1/2 · 9.8 m/s² · t²
8 m / 9.8 m/s² = t²
t = 0.9 s
Then:
3 m = v0x · 0.9s
3 m/ 0.9 s = v0x
v0x = 3.3 m/s
The speed at which the ball roll off the end of the table is 3.3 m/s
