Explanation:
Below is an attachment containing the solution.
Answer:
<h2>8.0995×10^-21 kgms^-1</h2>
Explanation:
Mass of proton :
![m_P=1.67\times 10^-^2^7\:kg\\](https://tex.z-dn.net/?f=m_P%3D1.67%5Ctimes%2010%5E-%5E2%5E7%5C%3Akg%5C%5C)
Speed of Proton:
![v_P=4.85\times 10^6](https://tex.z-dn.net/?f=v_P%3D4.85%5Ctimes%2010%5E6)
Linear Momentum of a particle having mass (m) and velocity (v) :
![-> p =m->v\:\:\: (1)](https://tex.z-dn.net/?f=-%3E%20p%20%3Dm-%3Ev%5C%3A%5C%3A%5C%3A%20%281%29)
Magnitude of momentum :
![p=mv\:\:\: (2)](https://tex.z-dn.net/?f=p%3Dmv%5C%3A%5C%3A%5C%3A%20%282%29)
Frome equation (2), magnitude of linear momentum of the proton :
![p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1](https://tex.z-dn.net/?f=p_P%3Dm_P%5C%3Av_P%5C%5C%5C%5Cp_P%3D1.67%5Ctimes%2010%5E-%5E2%5E7%20%5C%3Akg%5Ctimes4.85%5Ctimes%2010%5E6%5C%3Ams%5E-%5E1%5C%5C%5C%5Cp_P%3D%208.0995%5Ctimes%2010%5E-%5E2%5E1%5C%3Akgms%5E-%5E1)
Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity![=70 mph\approx 31.29 m/s](https://tex.z-dn.net/?f=%3D70%20mph%5Capprox%2031.29%20m%2Fs)
velocity after 5 s is ![30 mph\approx 13.41 m/s](https://tex.z-dn.net/?f=30%20mph%5Capprox%2013.41%20m%2Fs)
Therefore acceleration during these 5 s
![a=\frac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7D%7Bt%7D)
![a=\frac{13.41-31.29}{5}=-3.576 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B13.41-31.29%7D%7B5%7D%3D-3.576%20m%2Fs%5E2)
therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s
![0=31.29-3.576\times t](https://tex.z-dn.net/?f=0%3D31.29-3.576%5Ctimes%20t)
![t=\frac{31.29}{3.576}=8.75 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B31.29%7D%7B3.576%7D%3D8.75%20s)
(b)total distance traveled before stoppage
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![0^2-31.29^2=2\times (-3.576)\cdot s](https://tex.z-dn.net/?f=0%5E2-31.29%5E2%3D2%5Ctimes%20%28-3.576%29%5Ccdot%20s)
s=136.89 m