A shell is fired from the ground with an initial speed of 1.74 ✕ 103 m/s at an initial angle of 58° to the horizontal.
1 answer:
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Answer:
The speed of electron is and the speed of proton is 2468.02 m/s.
Explanation:
Given that,
Electric field, E = 560 N/C
To find,
The speed of each particle (electrons and proton) 46.0 ns after being released.
Solution,
For electron,
The electric force is given by :
Let v is the speed of electron. It can be calculated using first equation of motion as :
u = 0 (at rest)
For proton,
The electric force is given by :
Let v is the speed of electron. It can be calculated using first equation of motion as :
u = 0 (at rest)
So, the speed of electron is and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.
Answer:
E= 50.1*10³ N/C
Explanation:
Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:
We know that v₀ = 0 (it starts from rest), that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:
According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.
Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:
F = q*E (1)
In this case, q=e= 1.6*10⁻19 C
But this force, can be expressed in this way, according Newton's 2nd Law:
F = m*a (2) ,
where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.
From (1) and (2), we can solve for E, as follows:
⇒ E = 50.1*10³ N/C