Answer:
No..
Explanation:
As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.
As the person is on circular arc constantly moving it will never meet that drop.
There are 3 significant figures, if that answers the question.
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>
Answer:
The gazelles top speed is 27.3 m/s.
Explanation:
Given that,
Acceleration = 4.2 m/s²
Time = 6.5 s
Suppose we need to find the gazelles top speed
The speed is equal to the product of acceleration and time.
We need to calculate the gazelles top speed
Using formula of speed
Where, v = speed
a = acceleration
t = time
Put the value into the formula
Hence, The gazelles top speed is 27.3 m/s.
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