Given:
object = 20kg
terminal speed of object = 80 m/s
According to the problem, drag force is proportional to speed, so Fd = kv ; k is some constant
At terminal velocity Vt: Fg = Fdmg = kVtk = mg / Vt = (20.0)(9.8)/(80.0) = 2.45 kg/s
<span>Fd = kv = 2.45v</span>
Fd = 2.45 (30.0) = 73.5 N
Answer:
X = 2146.05 m
Explanation:
We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.
We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:
X = Vt (1)
However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:
t = √2h/g (2)
Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:
t = √(2*2500)/9.8
t = 22.59 s
Now replacing into (1) we have:
X = 95 * 22.59
<h2>
X = 2146.05 m</h2>
This is the distance where the package should be released.
Hope this helps
A line that is falling towards the x axis represents an object that is negatively accelerating, or slowing down. When the line hits the x axis, the object has stopped moving. If the graph continues below the x axis, the object has changed direction and is moving backwards at increasing velocity.
So if the formula for work is force times displacement times cosine(theta), you'd plug in the numbers
100x5 (since there's no angle in the problem, cosine(theta) isn't used
100x5 = 500
So the answer would be B.
Hope that helps!
