Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F =
q =
we calculate
q =
q = Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer:
The kinetic energy of the system after the collision is 9 J.
Explanation:
It is given that,
Mass of object 1, m₁ = 3 kg
Speed of object 1, v₁ = 2 m/s
Mass of object 2, m₂ = 6 kg
Speed of object 2, v₂ = -1 m/s (it is moving in left)
Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,
E = 9 J
So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.
Taking specific heat of lead as 0.128 J/gK = c
We have energy of ball at 7.00 meter height = mgh =
When leads gets heated by a temperature ΔT energy needed = mcΔT
= ΔT
Comparing both the equations
=
ΔT
ΔT = 0.536 K
Change in temperature same in degree and kelvin scale
So ΔT = 0.536