Answer
given,
time interval = 11.3 s
a) initial velocity, vi = 15 m/s
final velocity, v_f = -5.30 m/s
a = -1.79 m/s²
the direction is along left side
b) initial velocity, vi = -5.30 m/s
final velocity, v_f = -15 m/s
a = -0.858 m/s²
the direction is along left side
c) initial velocity, vi = 15 m/s
final velocity, v_f = -15 m/s
a = -2.65 m/s²
the direction is along left side
Answer:
The source is at a distance of 4.56 m from the first point.
Solution:
As per the question:
Separation distance between the points, d = 11.0 m
Sound level at the first point, L = 66.40 dB
Sound level at the second point, L'= 55.74 dB
Now,
where
I = Intensity of sound
Now,
Similarly,
Now,
Solving the above quadratic eqn, we get:
R = 4.56 m
Answer:
26.64 m
Explanation:
Given the following :
Acceleration at ocean surface = 0.0800 m/s²
Distance covered if initial speed = 0.700 m/s and accelerates to a speed of 2.18m/s
Using the equation :
v² = u² + 2as
Where ;
v = final velocity ; u = initial velocity ; a = acceleration ; s = distance covered
Therefore,
v² = u² + 2as
2.18² = 0.7² + (2 × 0.08 × s)
4.7524 = 0.49 + 0.16s
4.7524 - 0.49 = 0.16s
4.2624 = 0.16s
s = 4.2624 / 0.16
s = 26.64 m
