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Explanation:
Let the east direction to be i^ and north direction be j^.
Thus displacement of the man S=8i^+6j^
So, magnitude of displacement ∣S∣=62+82=100=10 m
The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:
h(t)=g/2(1.5)²+14.7(1.5)
=-4.9 x 2.25 + 22.05
=11.025m
The ball then falls for 49+11.025=60.025m, which takes:
g/2t²=60.025
t²=12.25
t=3.5 secs
Total time: 1.5+3.5=5 seconds
Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s
A. 0.5kg
To get this answer you need to follow the equation of KE=0.5*mv^2
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a.
As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.
That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J.
Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
