The kenitc energy of the bullet lowers as it keeps going up.
Because gravity is pushing the bullet down as the bullet goes up.
I'm pretty sure that the way to put this answer.
Answer:
Explanation:
F = ma and
We have F, we have m, but in order to solve for v, we need a.
30.0 = 3.00a so
a = 10.0 m/s/s. Plug that in for a in the second equation and solve for v:
so
v = 10.0(3.00) so
v = 30.0 m/s
Friction can be bad by being too strong or too weak.
<span>Sometimes, when it is too strong, it decreases efficiency since some energy is wasted and turns to heat. Friction can also d</span><span>amage equipment/objects like when you slide it on the floor.
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When friction is too weak, like for instance when there is black ice- our center of gravity is displaced too quickly and we can fall. Likewise, if there is a lot of slush on the ground, cars can slip and slide.
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
