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Svet_ta [14]
3 years ago
6

Captain hook orders peter pan to walk the plank from which he falls into the sea 18m below. Peter pan claims that his life flash

ed before his eyes during his fall. How much time is available for this?
Physics
2 answers:
Neporo4naja [7]3 years ago
6 0
You would use
d = v0*t + 0.5at^2
Solve for t. v0 is 0, d is 18 and a is -9.81

t = sqrt(18/4.9)
Nat2105 [25]3 years ago
3 0
The distance fallen through gravity in ' t ' seconds is

        Distance  =  (1/2) (acceleration of gravity) (t²)

           18 meters  =  (1/2) · (9.8 m/s²) · (t²)

Divide each side by (4.9 m/s²):

                           (18 m) / (4.9 m/s²)  =  t²

                                 3.67 sec²  =  t²

Square root each side:    1.92 sec  =  t  .    (rounded) 

That's not much time.  But Peter Pan never got old, so
presumably he hasn't got too much life to flash through. 
He may be OK with 1.92 seconds.  Maybe even enough
time to linger momentarily over some of the juicier parts.
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Astrology is the answer

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3 years ago
8. A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation f
pychu [463]

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

7 0
2 years ago
A"car"initially"at"rest"experiences"a" constant"acceleration"along"a"horizontal" road."the"position"of"the"car"at"several" succe
marysya [2.9K]

In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration.  So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.

7 0
3 years ago
Ok, so this question is probably really easy but I can't really be bothered to answer it, terrible I know, but I thank all usefu
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Without a bulb energy cant go through and it would be an open circuit blocking the energy from coming out.
3 0
3 years ago
A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c
marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
  • Distance=s=96m
  • Final velocity=v

Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

\\ \Large\sf\longmapsto v^2=576+768

\\ \Large\sf\longmapsto v^2=1344

\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

3 0
3 years ago
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