What is the stone’s velocity after 1 second? m/s
2 answers:
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To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by
Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So
x = Displacement
t = time
With the data we have we can find the time as well
With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,
Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
Answer:
F = 3.86 x 10⁻⁶ N
Explanation:
First, we will find the distance between the two particles:
where,
r = distance between the particles = ?
(x₁, y₁, z₁) = (2, 5, 1)
(x₂, y₂, z₂) = (3, 2, 3)
Therefore,
Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:
where,
F = magnitude of force = ?
k = Coulomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 2 x 10⁻⁸ C
q₂ = magnitude of second charge = 3 x 10⁻⁷ C
r = distance between the charges = 3.741 m
Therefore,
<u>F = 3.86 x 10⁻⁶ N</u>