Physics Questions & Answer Key

Question:

Mr. Physics drops a rock down a 250-meter deep well. Upon impact with the bottom of the well a short burst of sound is produced with a frequency of 250 Hz. How much time after he drops the rock does he hear it hit the bottom of the well.


Answer:

d = 250 m

f = 250 Hz

v = 343 m/s

u = 0 m/s

d = ut + ½ gt²

250 = (0)(t) + ½(9.8)( t²⁾

t = 7.143 sec  (going down well)

d = vt

250 = (343)(t)

t = 0.729 sec (going back up well)

t = 7.143 + 0.729 = 7.872 sec

  

Physics Question & Answer Key

Question:

Two identical ice skaters are holding onto opposite ends of a very long spring with 

a spring constant of 650 N/m. While spinning in a circle, the skaters frequency of 

revolution is measured to be 0.85 Hz. Calculate the mass of the skaters. 


Answer:

f = (1/2 π)(√ k/m)

0.85 = (1/2 π)(√ 650/m)  

m = 22.8 kg 

22.8 x 2 = 45.6 kg  (the mass need to double since circular motion is the resultant 

of two perpendicular SHM with the same frequency and amplitude and phase 

difference of π/2.)

Physics Question & Answer Key

Question:

A car traveling at constant speed negotiates a frictionless banked curved. Calculate the speed of the car when the road is banked at 20 degrees and the radius of its path is 50 meters.

Answer:

When the road is banked at angle 20 degrees,
The normal force is inclined at angle 20 degrees to the vertical
Its horizontal component
mg sin 20 = centripetal force = mv^2/r
Its vertical Component balances the weight 
mg cos 20 = mg 
Dividing the first by the second equation
tan 20 = v^2/(rg)
v^2 / 50 = (9.8)(tan 20)
v = 13.35 m/s

Physics Question & Answer Key

Question 1:

A 75 kg person escapes from a burning building by walking out a window 25m above a catching net. Assuming that air resistance is a constant 95N on the person during the fall, calculate the person's velocity just before hitting the net.

F = ma

(75)(9.8) - 95  = 75a

a = 8.53 

s = (1/2)(a)(t^2)

25 = (1/2)(8.53)(t^2)

t = 2.42

 v = u + (a)(t)
    = 0 + (8.5)(2.42)
    = 20.66 

 

Question 2:

The maximum force a grocery sack can withstand and not rip is 250 N. Calculate the greatest acceleration 20kg of groceries can be lifted from the floor to the table such that the sack does not break?

Fg = mg

     = (20)(9.8) = 196

F = ma

250 - 196 = (20)(a)

 a = 2.7 

   

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