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mote1985 [20]
3 years ago
15

Measure Your Reaction Time Here's something you can try at home-an experiment to measure your reaction time. Have a friend hold

a ruler by one end, letting the other end hang down vertically. At the lower end, hold your thumb and index finger on either side of the ruler, ready to grip it. Have your friend release the ruler without warning. Catch it as quickly as you can.If you catch the ruler 5.7 cm from the lower end, what is your reaction time?
Express your answer using two significant figures.
Physics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

Explanation:

I catch the ruler 5.7 cm from lower end that means my reaction time is equal to time of fall of ruler as free fall under gravity .

h = 1/2 gt²

t = \sqrt{\frac{2h}{g} }

= \sqrt{\frac{2\times 5.7}{9.8} }

= 1.078 s

= 1.1 s .

You might be interested in
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
Which astronomer supported the belief that earth was at the center fo the universe?
MissTica
B. Ptolemy believed that the earth was the center of the universe
4 0
4 years ago
Read 2 more answers
Which of the following items best embodies the physical property of conductivity?
Ostrovityanka [42]

Copper penny is the answer

8 0
3 years ago
Read 2 more answers
All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}

since v_y = 0 at the top of its trajectory. Plugging in the numbers,

t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}

8 0
3 years ago
Un avión de rescate en Alaska deja caer un paquete de provisiones a un grupo de exploradores extraviados. Si el avión viaja hori
posledela

Answer:

180.4 m

Explanation:

The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:    

$V_{ox}=v_x = 40 \ m/s$

   h = 100 m  

    x =?

     Height formula h:

     $h=g \times \frac{t^2}{2}$

      Time t is cleared:

     $t = \sqrt{\frac{2h}{g}}$

      $t = \sqrt{\frac{2 \times 100}{9.8}}$

      t = 4.51 sec

 Horizontal distance formula x:

       $x=V_x \times t$

        x = 40 m / sec x 4.51 sec

        x = 180.4 m

4 0
3 years ago
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