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Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
Explanation:
F = ma, and a = Δv / Δt.
F = m Δv / Δt
Given: m = 60 kg and Δv = -30 m/s.
a) Δt = 5.0 s
F = (60 kg) (-30 m/s) / (5.0 s)
F = -360 N
b) Δt = 0.50 s
F = (60 kg) (-30 m/s) / (0.50 s)
F = -3600 N
c) Δt = 0.05 s
F = (60 kg) (-30 m/s) / (0.05 s)
F = -36000 N
Ionic bonds occur when electrons are donated from one atom to another. Each type of atom forms a characteristic number of covalentbonds with other atoms. An example of that is a hydrogen atom with one electron in its outer shell forms only one bond, its out most orbital becomes filled with two electrons.
the outermost protons
