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

  

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 

   

Physics Question of The Week (5)

1. Electrons are ejected from a surface with speeds ranging up to 4.6 x 10⁵ m/s when light with a wavelength  λ = 625 nm is used.  

        a)    What is the work function of this surface?  

        b)    What is the threshold frequency for this surface?  

Answer Key for Physics Question of The Week (4)

F = qvB = mv² / r

F = (1.6 x 10^-19)(2 x 10⁷)(0.1)

   = 0.32 x 10^-12

r = [(1.67 x 10^-27)(2 x 10⁷)²] / 0.32 x 10^-12

   = 2.09 m

Physics Question of The Week (4)

A proton moves at a speed of 2 X 10⁷ m/s at right angles to a magnetic field with a magnitude of 0.10 T.  Find the radius of its circular orbit.  

Answer Key for Physics Question of The Week (3)

μ₀  =  ₄π x 10^-7  

I  =  5

r  =  30 cm =  .3 m

B  = 2 x 10^-5

B  =  (μ₀I) / (2πr)

(₄π x 10^-7)(5)  / (2π)(0.3)  =  (20 x 10^-7)  /  (0.6)  =  (10 x 10^-7)  /  (0.3)

(2 x 10^-5)(tan θ)  =  (10 x 10^-7)  /  (0.3) 

θ = 9.46⁰

Physics question of the week (3)

A boat is sailing due north, as indicated by its compass needle, in a location where Earth's magnetic field is 2.0 X 10^-5 T. The boat’s captain inadvertently places his radio on the shelf directly above the compass.  If the 5.0 A current carrying wire of the radio is aligned in a north-south direction and is 30 cm directly above the compass, how much from due north with the compass needle be deflected?  

Answer Key for Physics question of the week (2)

a. v = (3050)(-4) - (1750)(15) / (1300)
      = -30 m/s (or 30 m/s west)


b. p = (1750)(-4 - 15)
       = -33000 Ns


c. Inelastic  

Physics question of the week (2)

A 1750 kg truck is heading east at 15 m/s when it collides with a 1300 kg car traveling
west. After the collision the car and truck are stuck together and slide west at 4.0 m/s.

a)What was the initial velocity of the car?
b) What was the impulse felt by the truck?
c) Was this collision elastic or inelastic?

Answer key for physics question (1)

(a) R = v /I = 120 / 0.25 = 480 Ω


(b) P = IV = (0.25)(120) = 30 w


(c) I = q / t ; q = (0.25)(15) = 3.75 C


     (3.75)(1 / 1.6 x 10^-19) = 2.34 x 10^-19  e^-

Physics question of the week (1)

1.  A blender draws 0.25 A when connected to a 120 V outlet.
     a) What is the resistance of the blender?
     b) How much power does it use?
     c) How many electrons pass through the blender in 15 s?