Explanation:
As the given spheres are connected by a thin wire so, the potential on the spheres are the same.
......... (1)
Hence, total charge will be as follows.
= Q = -95.5 nC .......... (2)
Using the above two equations, the final equation will be as follows.
and,
Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.
=
= 82.714 nC
Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.
Answer:
50 N
Explanation:
Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:
The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.
The horizontal component of the forces:
F₁ + F₂ = -40N + F₂ = 0
F₂ = 40N
The vertical component of the forces:
F₁ + F₂ - mg = 0 + F₂ - mg = 0
F₂ = mg
If I assume the gravitational constant g = 10 m/s²:
F₂ = (3 kg) * (10 m/s²) = 30N
Adding the horizontal and vertical components of the force F₂:
F₂ = √((40N)² + (30N)²) = 50N
Given :
Initial speed of car A is 15 m/s and initial speed of car B is zero.
Final speed of car A is zero and final speed of car B is 10 m/s.
To Find :
What fraction of the initial kinetic energy is lost in the collision.
Solution :
Initial kinetic energy is :
Final kinetic energy is :
Now, fraction of initial kinetic energy loss is :
Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J