Answer:
Power output = 96.506 watts
Explanation:
Drag coefficient (Cd) = 0.9
V = 7.3 m/s
Air density (ρ) = 1.225 kg/m^(3)
Area (A) = 0.45 m^2
Let's find the drag force ;
Fd=(1/2)(Cd)(ρ)(A)(v^(2))
So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N
Drag power = Drag Force x Drag velocity.
Thus drag power, = 13.22 x 7.3 = 96.506 watts
Well this question looks like it makes some assumptions. So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power". The reason is that for a given race the cars must overcome losses associated with motion. Since they all travel the same distance, the amount of work will be the same for both. This is because work is force times distance. If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done. So for a slower car, time t will be larger. The power ratio W/t will be smaller for the longer time (slower car).
Answer: A raindrop
Explanation:
It would be a raindrop depending on if the ship is moving or not. If the ship is not moving, then the raindrop would have more momentum. But, if the ship were moving, then it would have more momentum. Because momentum equals mass times velocity.
i am right their with you i do believe it is A or D but i am more certain of D
At the beginning of a basketball game, a referee tosses the ball straight
up with a speed of 4.6 m/s. a player cannot touch the ball until after it reaches its maximum height and begins to fall down. what is the minimum time that a player must wait before touching the ball? s
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kinemaic equation
v=u-at
0=4.6-9.81xt
t=4.6/9.81 ... about half a second
