Explanation:
According to newtons first law of motion:
'' a body will continue in its state of rest or uniform motion along a path unless it is acted upon by an external force".
A body in equilibrium that is floating will be stable and not move in any direction. Even if it moves, the motion will be constant wouldn't change.
- To move the body in any direction, one has to swim.
- Swimming is the application of an external force to counter the balanced forces at equilibrium on a body.
- This works when the net external force is greater than that the balanced forces.
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Answer: (a) t = 5.44 sec
(b) vf = 53.31 m/s
(c) s = 5.0m
Explanation: from the question, given data
the Height of the tower, h = 145m
from question
(a)
the initial velocity, v₁ = 0 m/s
s = v₁t + 1/2 gt²
-145 m = 0(t) + 1/2 (-9.8t²)
t² = 145/4.9
t² = 29.59
t = 5.44 sec
(b)
the speed of the sphere at the bottom of the tower is
vf² = vi² +2as
vf² = 0 + 2(-9.8 × -145)
vf² = 2842
vf = 53.31 m/s
(c)
when caught, the sphere experiences a deceleration of;
a = -29.0g
the time it would take to decelerate becomes;
vf = vi + at
0 = (53.31) + (-29 ×9.8)t
where t = 53.31 / 284.2
t = 0.1876 sec
∴ the distance travelled during the deceleration becomes;
vf² = vi² + 2as
s = (vf² - vi²) / 2a
s = (0 - 53.31²) / 2×-29×9.8
s = -2841.9561 / -568.4
s = 4.99 ≈ 5.0m
i hope this helps, cheers
By collapsing and crumpling, the force and energy is absorbed by the crumple zone and it is not transferred to the passengers inside. This is a safety feature for the passengers.
Time = (distance) / (speed)
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Time = (450 km) / (100 m/s)
Time = (450,000 m) / (100 m/s)
Time = <em>4500 seconds </em>(that's 75 minutes)
Note:
This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.
If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.
The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.
To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:
h = -16t^2 + 127t
We substitute 55 seconds to t and obtain,
h = -16(55)^2 + 127(55)
h = - 41415
