The net force on q2 will be 1.35 N
A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.
Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.
We have to find the net force on q2
At first we will find Force due to q1
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²
F = 450 × 10⁻³
F₁ = 0.45 N (+)
Now we will find Force due to q2
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²
F = 1800 × 10⁻³
F₂ = 1.8 N (-)
So net force (F) will be
F = F₂ - F₁
F = 1.8 - 0.45
F = 1.35 N
Hence the net force on q2 will be 1.35 N
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Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
So now we can find the magnitude of the initial velocity:
The work of Brian mind interface is to learn that’s my guess lol
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Anyone can use it and </span>Scientific inquiry involves a person, research tools, space, writing pad, and one, etc.
