The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:
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a) Vertical velocity: 5.9 m/s
b) Horizontal velocity: 5 m/s
Explanation:
a)
The motion of the ball is the motion of a projectile, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction)
- A uniformly accelerated motion (constant acceleration) along the vertical direction
Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:
where:
is the vertical velocity at time t
is the initial vertical velocity (zero because the ball has been thrown horizontally)
is the acceleration of gravity (here we take downward as positive direction)
Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:
b)
The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.
However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.
As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.
Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still