An object is falling downward at a rate of 25 m/s. Two seconds later, what is its acceleration?
1 answer:
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Answer:
6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.
Explanation:
What matters is the part , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:
For both cases, + and -, after a certain time (
), the displacement <em>y</em> of the wave will be determined by the
term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply
.
To know which side, right or left of the origin, would go through the origin after a time (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation
(we would be looking where in space is what we would have in the future in time). The term would be then
, which at the origin is
. This would mean that, when the original equation has
, we must have that
for
to be equal to
, and when the original equation has
, we must have that
for
to be equal to
<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>
<em />
In conclusion, when , the part of the wave on the positive side (
) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
W = K.E
Therefore, the ball traveled 0.827 m