The two elements are in the same period, with Element R the first element in the period and Element Q the last element.
Gravitational potential energy=mass*gravitational acceleration*height
Kinetic energy = 0.5*mass*velocity²
So with the given data
K.E 0.5*1*x²=12.5 v²=12.5÷(0.5*1)
v=√12.5÷(0.5*1) v=5
GPE mass*gravitational acceleration*height
1*9.81*h=98
h=98÷(9.81*1)
h= 9.98 m
<em>Kinetic (mechanical) energy</em>
Answer:
A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.
B. Veloctiy (Vb) = 1.66m/s
Explanation:
Given the following data
x(a) = 0.3m
x(b) = 0
q = 1.6×10^-19
Q = 24nc
r = 0.15m
Required: the motion of the electron and the velocity (Vb)
1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B
2. Potential energy and kinetic energy are given by
U(a) + K(a) = U(b) + K(b). . .1
Initial P.E and K.E are given as
U(a) = kQ/√x²(a) + a2
By substitution, we have
U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)
U(a) = -1.03×10^-16
Final P.E and K.E are given as
U(b) = KQ/√x²(b) + a2
By substitution, we have
U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²
U(b) = -2.3×10^16
3. By substitution into equation 1 becomes
-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2
V(b) = √2×1.27×10^-16/9.1×10^31
V(b) = 1.66×10^7m/s
