Ricks velocity would be zooomin out because it would fall off so strongly so it’d change and it’s weight too
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The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
Loose 10 pounds im pretty sure
<span>anwser will be
F = ma
where
F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet
Substituting appropriately,
F = 0.005a --- call this Equation 1
Next working equation is
Vf^2 - Vo^2 = 2as
where
Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel
Substituting appropriately,
326^2 - 0 = 2(a)(0.83)
a = 64,022 m/sec^2
the anwser will be
Substituting this into Equation 1,
F = 0.005(64,022)
F =320.11 Newtons
Hope this helps. </span><span>
</span>
Since static friction is the minimum force required to just start the motion of a stationary object.
Here if we need to start an object from rest then we required F = 700 N
So for the first part of the above problem Force will be F = 700 N
Now if the box is already moving then we will have to use kinetic friction force between box and floor
now we can write the equation of net force as
here
now we will have
