2. The mathematician john von Neumann used the term micro century to denote the

Question

2. The mathematician john von Neumann used the term micro century to denote the optimum length of a lecture and according to his definition One micro century s 52 minutes and 35.69 seconds. Assuming one PHYS 2321 class is 1 hour 15 minutes, How many micro centuries of PHYS 2321 classes will you take during summer 2015 semester in total? 5. A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5 m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.4 m/s^2 until he manages to catch his friend. a) How much time does it take for the bicyclist to catch his friend? b) How far does he have to travel to catch his friend? c) What is the bicyclist?s speed when he catches his friend? 6. A simple way to test reaction times is to have someone drop a ruler in between your two fingers and you catch it as quickly as you can. a) If the ruler starts at one end (0 cm). and you catch it with your fingers on the 7.5 cm mark, what is your reaction time? b) If the best human reaction times are around 0.10 s. would you be willing to set up a game using the same experiment with a $100 dollar bill, which is about 15 cm in length? The point of the game is you hold the $100 dollar bill while the player has their fingers at the exact middle of the bill. You then drop It, and if the player manages to catch the $100 bill, they get to keep it...if they don?t, they owe you $5. 7. Standing side by side, you and a friend step off a bridge at different times and fall for 1.6 s to the water below. Your friend goes first and you soon follow after your friend has dropped a distance of 2.0 m. a) When your friend hits the water, are you two still separated by 2.0 m? Is it less? Is it more? (No calculation allowed!) b) Verify your answer to a) by calculating the actual separation distance.

Explanation / Answer

5)

a)the positon of the cyclist with the briken bike is

x1=u1+1/2at^2

u1=0

x1=1/2*(2.4)t^2

after 2 second postion of friend is at

x2=vt

x2=3.5*2

x2=7m

after that postion of a friend is at

x2=7+(3.5)t

putting x1 and x2 together we get

1/2*(2.4)t^2=7+(3.5)t

(1.2)t^2-3.5t-7=0

solving the quadratic equation we get

t=-1.36 or 4.28 sec

ignoring the negative value

t=4.28 sec---answer to a

b)X1=1/2*(2.4)*(4.28)^2

X1=22m

c)v1=v0+at

v0=0

so v1=0+(2.4)*(4.28)

v1=10.7m/sec

please raise diferent questions for other questions

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