Physics Question & Answer Key

A proton with an initial horizontal velocity of 2.3X10^6 m/s enters a region of uniform electric field strength of 1.1X10^5 N/C between two oppositely charge parallel plates. If the plates are 1.3cm apart and 6.7cm long, how far below the top plate must the particle be shot so as to just miss hitting the right edge of the top plate? Ignore gravitational forces.

t = d / v

   = 0.067 / (2.3e6)

   = 2.9e-8

E = F / q

F = Eq

   = (1.1e5)(1.6e-19)

   = 1.76e-14

F = ma

a = F / m

    = (1.76e-14) / (1.67e-27)

    = 1.05e13

s = ut + .5at²

   = 0 + .5(1.05e13)(2.9e-8)²

   = 0.0045 m (from top)

Physics Question & Answer Key

A 5.00 x 10^ -2 kg mass with a charge of .75μC is hung by a thin insulated thread. A charge of -0.9μC is held .15m directly to the right so that the thread makes an angle Θ with the vertical. What is the angle Θ and the tension in the string?

Force between charges:

F = [(9e9)(0.75e-6)(0.9e-6)] / (0.15²) 

   = 0.27 N

Gravitational force:

mg = 5e-2 x 9.8 

      = 0.49 N

From a force vector diagram:

0.49N vertically down & 0.27N horizontal - -  combining to form θ from vertical) 

tan θ = 0.27 / 0.49 

θ = 28.9º

sin θ = 0.27 / T 

T =0.27 / (sin28.9) 

   = 0.56 N

Physics Question & Answer Key

Question:

When two organ pipes open at both ends are sounded at their first harmonic, 4 beats/second are heard. One pipe is 1000 mm long. Calculate the length of the other pipe.

Answer:

V = 343 m/s

L₁ = 1000 mm = 1 m

λ = 2L₁ = 2 m

f_beat = ½f₁ – f₂½

4 = (343 / 2) – (343 / l₂)

λ₂ = 2.048

λ₂ = 2L₂

2.048 = 2 L₂

L₂ = 1.024 m = 1024 mm

Physics Question & Answer Key

Question:

A vibrating string on a violin is 330 mm long and has a fundamental frequency of 659 Hz. Tge string is then pressed on against the fingerboard at a point of 60 mm from its end.
a) Calculate the fundamental frequency when the string is pressed against the fingerboard. 
b) Calculate the wavelength of the sound wave produced. 
c) Calculate the frequency of the next two harmonics. 


Answer:

f = 659 Hz
L = 0.33 m
L_{1 = 0.33 - 0.06 = 0.27 m}

a) 

v = λf
   = (659)(2)(0.33)
   = 434.94 m/s

f_{1= (434.94) / (2)(0.27)}

   _{= 805 Hz}

_b) 

λ = v / f
   = (343) / (805)
   = 0.43 m

c)

f_{2 = (2)(434.94) / (2)(0.27)}

    _{= 1611 Hz}

f_{3 = (3)(434.94) / (2)(0.27)}

    _{= 2416 Hz}

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