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<h2>The work done , when body moves along the plane </h2>
Explanation:
A body is projected from bottom of the inclined plane . When body is going up the plane .
The downward force = m g sinθ is developed due to its weight
As body is moving upwards , the force of friction will act downwards
The force of friction = μ R
here μ is the coefficient of friction ans R is the normal reaction
Thus force of friction f = μ mg cosθ
Let the acceleration upwards is a
The upward force required = m a
Thus m a = mg sinθ + μ mg cosθ
or acceleration a = g ( sinθ + μ cosθ )
The work done in moving upwards W = F S
Thus W = mg ( sinθ + μ cosθ ) S
here S is the displacement on the plane
When body moves down , the force of friction acts upwards
Thus m a = m g ( sinθ - μ cosθ )
The work done W = m g ( sinθ - μ cosθ ) S
As the body is projected with velocity u
which can be calculated by the relation v² - u² = - 2 a X
Here v = 0 at the highest point
Thus u =
here a = g ( sinθ + μ cosθ )
Similarly , when it moves down , the initial velocity u = 0
Thus v² - 0 = 2 a x
or v =
here a = g ( sinθ - μ cosθ )
Answer:
The equilibrium position for the third charge is 69.28 cm
Explanation:
Given;
q₁ = -5.00 x 10⁻⁹ C
q₂ = -2.00 x 10⁻⁹ C
q₃ = 15.00 x 10⁻⁹ C
distance between q₁ and q₂ = 40.0 cm = 0.4 m
(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)
At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃
According to Coulomb's law, repulsive or attractive force between charges is calculated as;
where;
F is repulsive or attractive force between charges
K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²
r₁ is the distance between q₁ and q₂
q₁, q₂ and q₃ are the charge
distance between q₂ and q₃, r₂ is calculated as;
Therefore, the equilibrium position for the third charge is 69.28 cm