The weight of a standard object defined as having a mass of exactly 2.9 kg is measured to be 28.449 n. in the same laboratory, a
2 answers:
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Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
<h3>Given, </h3>
Force,F = 4000 N
Area,a = 50 m²
<h3>We know that, </h3>Pressure = Force/Area
★ Putting the values in the above formula,we get:
Answer:
12 m
Explanation:
The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:
where
s is the distance
u is the initial velocity
t is the time
a is the acceleration
For this problem,
and
u = 0, since we are considering the first second of motion
So, substituting t = 1 s, we find