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Dafna11 [192]
3 years ago
5

You are doing an experiment to determine your reaction time. Your friend holds a ruler. You place your fingers near the sides of

the lower part of the ruler without touching it. The friend drops the ruler without warning you. You catch the ruler after it falls 11.3 cm . What was your reaction time? please show formulas and explanation step by step
Physics
2 answers:
olga2289 [7]3 years ago
4 0
I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
    increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.

Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .

Here's the formula for the distance an object falls from rest
in a certain time:

                Distance = (1/2) (gravity) (time)²

On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...

                              11.2 cm  =  (1/2) (9.8 m/s²) (time)²
or
                         0.112 meter  =  (4.9 m/s²) (time)²                      

Divide each side
by  4.9 m/s² :        (0.112 m) / (4.9 m/s²)  =  time²

                            (0.112 / 4.9)  sec²  =  time²

Square root
each side:            time = √(0.112/4.9  sec²)

                                  =  √ 0.5488 sec²

                                  =      0.74 second     (rounded)  

Nataliya [291]3 years ago
4 0

Answer:

0.152 seconds

Explanation:

Distance travelled by the ruler = 11.3 cm = s

Initial velocity of the ruler = 0 = u

Final velocity of the ruler = v

Time taken by you to catch the ruler = t

a = Acceleration due to gravity = 9.81 m/s²

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow 0.113=0t+\frac{1}{2}9.81t^2\\\Rightarrow t^2=\frac{2\times 0.113}{9.81}\\\Rightarrow t=\sqrt{\frac{2\times 0.113}{9.81}}\\\Rightarrow t=0.152\ s

So, it took you 0.152 seconds to react to the ruler falling.

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You are designing an airplane propeller that is to turn at 2400 rpm (Fig. 9.13a). The forward airspeed of the plane is to be 75.
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5 0
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An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equi
tino4ka555 [31]

Answer: An archer pulls her bowstring back 0.400m by exerting a force that increases uniformly from zero to 230N. The equivalent spring constant of the bow will be 575N/m

Explanation: To find the answer, we have to know about the Simple Harmonic Motion.

<h3>What is Simple Harmonic Motion?</h3>
  • A particle is said to execute simple harmonic motion, if it moves to and fro about mean position under the action of restoring force.
  • The restoring force is directly proportional to its displacement from the mean position and always directed towards the mean position.
  • If x is the displacement from the mean position, and F is the restoring force, then

                   F∝-x

                   F=-kx where, k is called the spring constant.

<h3>How to approach the problem?</h3>
  • Given that, an archer pulls her bowstring back 0.400m by exerting a force that increasing from zero to 230N.
  • Here, from the question given, we can write, x=-0.400m and F=230N.
  • Thus, our spring constant k will be,

                       k=-F/x= (230)/ (0.400) =575 N/m.

Thus, we can conclude that, the equivalent spring constant of the bow will be 575N/m.

Learn more about the Simple Harmonic Motion here:

brainly.com/question/28019840

#SPJ4

<h3 />

8 0
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