The answer would be B, 19 protons. Hope this helps!
Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s
Answer:
upthrust or BUOYANT FORCE =Vdg
volume=LWH
upthrust=(4cm×5cm×2cm)×1g/cm²×g
upthrust=40cm³×1g/cm³×g
upthrust=40gf or 0.04kg×10m/s²=0.4N
weight of the displaced liquid is upthrust.
so mass=40g or 0.04kg
upthrust=40gf or 0.4Nand mass of the displaced liquid=40g or 0.04kg
please mark brainliest, hope it helped