Answer:
A = m³/s³ = [L]³/[T]³ = [L³T⁻³]
B = m³s = [L³T]
Explanation:
We have the equation:
V = At³ + B/t
where, the dimensions of each variable are as follows:
V = m³ = [L]³
t = s = [T]
substituting these in equation, we get:
m³ = A(s)³ + B/s
for the homogeneity of the equation:
A(s)³ = m³
<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>
Also,
B/s = m³
<u>B = m³s = [L³T]</u>
Answer:
Traction
Explanation:
Enter a curve slower than the posted speed if your vehicle has a high center of gravity or if surface Traction is less than ideal.
Traction is an act of drawing or pulling something over a surface specially a road or a track or it also defined as the grip of a tire on road or a wheel on rail. So speed when entering a curve should be slower when the tire has low traction to avoid accident.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

And the distance traveled downwards is:

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

Replacing into the first equation:

Rationalizing:

Let's call v1 the final speed of the package dropped from a height H. Thus:

Let v2 be the final speed of the package dropped from a height 4H. Thus:

Taking out the square root of 4:

Dividing v2/v1 we can compare the final speeds:

Simplifying:

The final speed of the second package is twice as much as the final speed of the first package.