Answer:
(a). The work done is 7001 MeV.
(b). The momentum of this proton is
.
Explanation:
Given that,
Speed = 0.993 c
We need to calculate the work done
Using work energy theorem
The work done is equal to the kinetic energy relative to the proton
![W=K.E](https://tex.z-dn.net/?f=W%3DK.E)
![W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7Bm_%7Bp%7Dc%5E2%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7D-m_%7Bp%7Dc%5E2)
Put the value into the formula
![W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2%7D%7B%5Csqrt%7B1-%28%5Cdfrac%7B0.993c%7D%7Bc%7D%29%5E2%7D%7D-1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2)
![W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2%7D%7B%5Csqrt%7B1-%280.993%29%5E2%7D%7D-1.67%5Ctimes10%5E%7B-27%7D%5Ctimes%283%5Ctimes10%5E%7B8%7D%29%5E2)
![W=1.122\times10^{-9}\ J](https://tex.z-dn.net/?f=W%3D1.122%5Ctimes10%5E%7B-9%7D%5C%20J)
![W=7001\ MeV](https://tex.z-dn.net/?f=W%3D7001%5C%20MeV)
(b). We need to calculate the momentum of this proton
Using formula of momentum
![p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7Bm_%7B0%7Dv%7D%7B%5Csqrt%7B1-%5Cdfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7D)
Put the value into the formula
![p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes0.993c%7D%7B%5Csqrt%7B1-%28%5Cdfrac%7B0.993c%7D%7Bc%7D%29%5E2%7D%7D)
![p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B1.67%5Ctimes10%5E%7B-27%7D%5Ctimes0.993c%7D%7B%5Csqrt%7B1-%280.993%29%5E2%7D%7D)
![p=1.404\times10^{-26}c](https://tex.z-dn.net/?f=p%3D1.404%5Ctimes10%5E%7B-26%7Dc)
![p=4.20\times10^{8}\ kg-m/s](https://tex.z-dn.net/?f=p%3D4.20%5Ctimes10%5E%7B8%7D%5C%20kg-m%2Fs)
Hence, (a). The work done is 7001 MeV.
(b). The momentum of this proton is
.
Answer:
![I_d = -3.454*10^{-10} \ \ (sin ( \ 170 \ t))](https://tex.z-dn.net/?f=I_d%20%3D%20-3.454%2A10%5E%7B-10%7D%20%5C%20%5C%20%28sin%20%28%20%5C%20170%20%5C%20t%29%29)
Explanation:
The displacement current
is given by the expression;
![I_d = \epsilon _o \frac{d \phi _ E }{dt}](https://tex.z-dn.net/?f=I_d%20%3D%20%5Cepsilon%20_o%20%5Cfrac%7Bd%20%5Cphi%20_%20E%20%7D%7Bdt%7D)
where
![\phi _ E = A.E \\ \\ \\\phi _E = A. \frac{V}{d} \\ \\ \\\phi _E = \frac{A}{d}(14*10^{-3})(cos \ 170 \ t ) \\ \\ \\\phi _E = \frac{820*10^{-4}}{5*10^{-3}} * (14*10^{-3})(cos \ 170 \ t ) \\ \\ \\\phi _E = 2296*10^{-4} (cos \ 170 \ t)](https://tex.z-dn.net/?f=%5Cphi%20_%20E%20%3D%20A.E%20%5C%5C%20%5C%5C%20%5C%5C%5Cphi%20_E%20%3D%20A.%20%5Cfrac%7BV%7D%7Bd%7D%20%5C%5C%20%5C%5C%20%5C%5C%5Cphi%20_E%20%3D%20%5Cfrac%7BA%7D%7Bd%7D%2814%2A10%5E%7B-3%7D%29%28cos%20%5C%20170%20%5C%20t%20%29%20%5C%5C%20%5C%5C%20%5C%5C%5Cphi%20_E%20%3D%20%5Cfrac%7B820%2A10%5E%7B-4%7D%7D%7B5%2A10%5E%7B-3%7D%7D%20%2A%20%2814%2A10%5E%7B-3%7D%29%28cos%20%5C%20170%20%5C%20t%20%29%20%5C%5C%20%5C%5C%20%5C%5C%5Cphi%20_E%20%3D%202296%2A10%5E%7B-4%7D%20%28cos%20%5C%20170%20%5C%20t%29)
![I_d = \epsilon _o \frac{d \phi _ E }{dt}](https://tex.z-dn.net/?f=I_d%20%3D%20%5Cepsilon%20_o%20%5Cfrac%7Bd%20%5Cphi%20_%20E%20%7D%7Bdt%7D)
![I_d = 8.85*10^{-12}(2296*10^{-4})(-170)(sin(170 \ t))](https://tex.z-dn.net/?f=I_d%20%3D%208.85%2A10%5E%7B-12%7D%282296%2A10%5E%7B-4%7D%29%28-170%29%28sin%28170%20%5C%20t%29%29)
![I_d = -3.454*10^{-10} \ \ (sin ( \ 170 \ t))](https://tex.z-dn.net/?f=I_d%20%3D%20-3.454%2A10%5E%7B-10%7D%20%5C%20%5C%20%28sin%20%20%28%20%5C%20170%20%5C%20t%29%29)
Answer: A constellations
Explanation: hope it helps
Answer:
A The force of air resistance becomes equal in magnitude to his weight.
Explanation:
When Carlos jumped out of the helicopter, he fell with a downward force which is proportional to his weight.
His weight is the product of his mass and acceleration due to gravity.
When the parachute is deployed, air resistance between the air molecules and the surface area of the parachute generates a drag force, which acts upwards.
If the drag forces exceeds his weight, then his net force will be directed upwards.
If his weight exceeds the drag forces, then the net force on his body will be downwards.
In the case that his drag force, and his weight are equal, there will be a zero net force on him, and he will experience no acceleration.
So, if after he opens his parachute, his acceleration due to gravity decreases to zero, then we can say that this is because the force of air resistance (drag) becomes equal in magnitude to his weight.