Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
<span>Earth (and hence the observer) moves.</span>
The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.
The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.
Answer:
Yes
Explanation:
There are two types of interference possible when two waves meet at the same point:
- Constructive interference: this occurs when the two waves meet in phase, i.e. the crest (or the compression, in case of a longitudinale wave) meets with the crest (compression) of the other wave. In such a case, the amplitude of the resultant wave is twice that of the original wave.
- Destructive interferece: this occurs when the two waves meet in anti-phase, i.e. the crest (or the compression, in case of a longitudinal wave) meets with the trough (rarefaction) of the other wave. In this case, the amplitude of the resultant wave is zero, since the amplitudes of the two waves cancel out.
In this problem, we have a situation where the compression of one wave meets with the compression of the second wave, so we have constructive interference.
