The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of
<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N
Then the work done by this force on the crate as it slides down the ramp is
<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J
The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so
<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2
Solve for <em>v</em> :
<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s
Virtual, upright, and larger than the object. :)
Checking a table of star sizes, we'll see that our Sun is a medium sized star, so the answer is D, average.
Answer:

Explanation:
We are going to use the accelerated motion formula
on the vertical component taking the upwards direction as positive.
Since <em>at maximum height the speed will be
</em>, the formula can be written as
, which for our values is:

Written to three decimal places.
When moving with the current, the speeds of the boat and current will add up. The first situation can then be represented by:
9 = 0.5(x + y)
While opposing the current, the speeds will be subtracted. Forming:
9 = 1.5(x - y)
The second option contains this equation system, so it is correct.