The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a
1 answer:
Answer: 247.67 V
Explanation:
Given
Potential At A
Potential at
when particle starts from A it reaches with velocity at Point while when it starts from C it reaches at point B with velocity
Suppose m is the mass of Particle
Change in Kinetic Energy of particle moving under the Potential From A to B
Change in Kinetic Energy of particle moving under the Potential From C to B
Divide 1 and 2 we get
on solving we get
You might be interested in
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’=
h’ =
h’ = 1/9 h
Answer:
90
Explanation:
The mass number of the missing daughter nuclei can be obtained as shown in the attached photo.
