Answer:
False
Explanation:
The second you let go its gonna release kinetic energy that's why it's potential
for this we apply, Heisenberg's uncertainty principle.
it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.
the formula is,
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
by rearranging,
Δx = h / 4π * m(e).Δv
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2
Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹
Δx = 0.011 X 10⁻³
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2
Δx = 6.63*10^-34 /2.05
Δx =3.23 X 10⁻³² m
therefore, we can say that the lower limits are 0.011 X 10⁻³ m for the electron and 3.23 X 10⁻³² m for the bullet
To know more about bullet problem,
brainly.com/question/21150302
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Answer:
8N
Explanation:
the body is subjected to pressure equals to force force inverse to the area of that body given that it is in motion
Answer:
v = ![\left[\begin{array}{c}0.66&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0.66%260%5Cend%7Barray%7D%5Cright%5D)
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:
The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:
