A 5.75 × 107 kg battle ship originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 550 m/s. show ans
1 answer:
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Answer:
2) 433 mph
Explanation:
The final velocity of the raindrop as it reaches the ground can be found by using the equation for a uniformly accelerated motion:
where
v is the final velocity
u = 0 is the initial velocity (the raindrop starts from rest)
a = g = 9.8 m/s^2 is the acceleration due to gravity
d = 2 km = 2000 m is the distance covered
Solving for v,
And keeping in mind that
1 mile = 1609 metres
1 hour = 3600 s
The speed converted into miles per hour is
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
where
is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
(c) 5550 m
The maximum playing time of the CD is
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
(d)
The angular acceleration of the CD is given by
where
is the final angular speed (when the CD is scanned at the outermost part)
is the initial angular speed (when the CD is scanned at the innermost part)
is the time elapsed
Substituting into the equation, we find