Explanation:
KE = 1/2*m*v^2
mass won't change
but velocity is
A: the force is pushing against the direction of the object, so velocity is decreasing, so KE is decreasing
B: the F net line is pretty much straight up, which means there is no net force pushing to the left, but velocity is still increasing
Why? Because if you use the pythagorean theorem, the hypotenuse is always longer than the legs.
Note: velocity only ever stays the same if there is no net force (or 0 acceleration), so all four of these are either decreasing or increasing
C: there is a net force to the right, so velocity is increasing
D: There is a net force to the right but velocity is to the left, so velocity is decreasing
Hope I got these right
The lunar lander landed on the moon
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
A medium of exchange<span> is an intermediary instrument used to facilitate the sale, purchase or trade of goods between parties. For an instrument to function as a </span>medium of exchange<span>, it must represent a standard of value accepted by all parties. In modern economies, the </span>medium of exchange<span> is currency.</span>
Okay assuming the log had no initial vertical motion (you threw it straight out), the distance to the bottom is
dist = 1/2 gt^2 = 1/2 x 9.8m/s/s x
(5.65s)^2 = 156.4m
the velocity at the bottom = g t = 9.m/s/s x 5.65s =55.4m/s
Hope this helps!
