Answer:
![KE=1.2036\times 10^{-12}\ J](https://tex.z-dn.net/?f=KE%3D1.2036%5Ctimes%2010%5E%7B-12%7D%5C%20J)
Explanation:
Given:
- charge on the alpha particle,
![q=2e=3.2\times 10^{-19}\ C](https://tex.z-dn.net/?f=q%3D2e%3D3.2%5Ctimes%2010%5E%7B-19%7D%5C%20C)
- mass of the alpha particle,
![m=6.64\times 10^{-27}\ kg](https://tex.z-dn.net/?f=m%3D6.64%5Ctimes%2010%5E%7B-27%7D%5C%20kg)
- strength of a uniform magnetic field,
![B=0.5\ T](https://tex.z-dn.net/?f=B%3D0.5%5C%20T)
- radius of the final orbit,
![r=0.5\ m](https://tex.z-dn.net/?f=r%3D0.5%5C%20m)
<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>
![q.v.B=m.\frac{v^2}{r}](https://tex.z-dn.net/?f=q.v.B%3Dm.%5Cfrac%7Bv%5E2%7D%7Br%7D)
![m.v=q.B.r](https://tex.z-dn.net/?f=m.v%3Dq.B.r)
where:
v = velocity of the alpha particle
![v=\frac{q.B.r}{m}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bq.B.r%7D%7Bm%7D)
![v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B3.2%5Ctimes%2010%5E%7B-19%7D%5Ctimes%200.5%5Ctimes%200.5%7D%7B6.64%5Ctimes%2010%5E%7B-27%7D%7D)
![v=1.2048\times 10^{7}\ m.s^{-1}](https://tex.z-dn.net/?f=v%3D1.2048%5Ctimes%2010%5E%7B7%7D%5C%20m.s%5E%7B-1%7D)
Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.
<u>We firstly find the relativistic mass as:</u>
![m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m](https://tex.z-dn.net/?f=m%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%20%7D%20%5Ctimes%20m)
![m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }](https://tex.z-dn.net/?f=m%27%3D%5Cfrac%7B6.64%5Ctimes%2010%5E%7B-27%7D%7D%7B%5Csqrt%7B1-%5Cfrac%7B%281.2048%5Ctimes%2010%5E7%29%5E2%7D%7B%283%5Ctimes%2010%5E8%29%5E2%7D%20%7D%20%7D)
![m'=6.6533\times10^{-27}\ kg](https://tex.z-dn.net/?f=m%27%3D6.6533%5Ctimes10%5E%7B-27%7D%5C%20kg)
now kinetic energy:
![KE=m'.c-m.c](https://tex.z-dn.net/?f=KE%3Dm%27.c-m.c)
![KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2](https://tex.z-dn.net/?f=KE%3D6.6533%5Ctimes%2010%5E%7B-27%7D%5Ctimes%20%283%5Ctimes%2010%5E8%29%5E2-6.64%5Ctimes%2010%5E%7B-27%7D%5Ctimes%20%283%5Ctimes%2010%5E8%29%5E2)
![KE=1.2036\times 10^{-12}\ J](https://tex.z-dn.net/?f=KE%3D1.2036%5Ctimes%2010%5E%7B-12%7D%5C%20J)
Young people
Explanation:
cuz old people can't do sports
Explanation:
The height of an object thrown upward from the floor of a canyon 106 ft deep, with an initial velocity of 120 ft per second. The equation is given by :
![h=-16x^2+120x-106](https://tex.z-dn.net/?f=h%3D-16x%5E2%2B120x-106)
Since, the depth of the canyon is (-106 feet) and the time taken by the object to rise to the height of the canyon wall is calculated as :
h = 0
On solving the above quadratic equation,
x₁ = 1.023 seconds
and
x₂ = 6.477 seconds
So, the time taken by the object to rise to the height of the canyon wall is 1.023 seconds (ignoring 6.477 seconds). Hence, this is the required solution.
Answer:
Well I'm going to go with A.
Explanation:
As per the question the mass of the boy is 40 kg.
The boy sits on a chair.
We are asked to calculate the force exerted by the boy on the chair at sea level.
The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.
At sea level the acceleration due to gravity g = 9.8 m/s^2
Therefore, the weight of the boy [m is the mass of the body]
we have m = 40 kg.
Therefore, w = 40 kg ×9.8 m/s^2
=392 N kg m/s^2
= 392 N
392 is not an option but I'm guessing you can round down 2 .to option A. 390...?
Most likely by moving an object and by moving it from one place to another and how fast and slow it goes to its destination.