Answer:
If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.
Explanation:
By momentum conservation, <em>and given the bullit and the recoil are in a straight line</em>, the momentum analysis will be <em>unidimentional</em>. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so
now we clear 
and use the given data to get that
<em>But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s</em>, then <u>if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained</u>, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity 
too.
Finally, using we calculate the kinetic energy as
230 Newton
Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
F = k(q1 q2)/ r^2
Distance between protons = d = 10⁻¹⁵ m
charge of proton = q = 1.6 × 10⁻¹⁹ C
Here q1=q2
electric force = F =230N
Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.
2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.
Find more about electric force of repulsion between nuclear protons
brainly.com/question/8404637
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Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
The momentum change =mass*velocity change. But sincevelocity change is not known another strategy must be used to find the momentum change. The strategy involves first finding the impulse (F*t = 1.0 N*s). Since impulse = momentum change, the answer is 1.0 N*s.
