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PHYSICS TEST PRACTICE
School's out forever,
School's out for summer,
School's out with fever,
School's out completely.
--- A. Cooper
Some Potentially useful constants:
Wavelength of peak sensitivity in human vision = 555nm
Coulomb’s Law constant k = 8.9875517873681764*109 N·m2 / C2
Permittivity of free space 0 = 8.85418782*10-12 C2 / N·m2
Permeability of free space 0 = 4*10-7 T·m / A
electron charge: e = 1.60217662*10-19 C
mass of an electron: me = 9.109383*10-31 kg
mass of a proton: mp ≈ 1.6726219*10-27 kg
mass of a neutron: mn ≈ 1.6749274*10-27 kg
Acceleration due to gravity g = 9.80665 m / s2
Newtons Gravitational constant G = 6.67408 ×10−11 N⋅m2/kg2
Planck’s constant h = 6.626070040*10-34 Js
Speed of light in vacuum c = 2.99792458*108 m/s
Note that for small angles sin(radians ) tan(radians ) radians
The threshold of hearing Io =10-12 W/m2
A sphere=4r2 & A circle=r2
One Year
365.24 days
Mechanics Formulas:
v f = vi + at
f = i + t
x f = xi + vi t + 12 at 2
f = i + i t + 12 t 2
F = ma
= I
PHYSICS TEST PRACTICE
1 1 1
h ' −i
= +
m= =
f i o
h o
1) Electromagnetic waves (multiple Choice)
Which of these are electromagnetic waves?
a. visible light
b. TV signals
c. cosmic rays
d. Radio signals
e. Microwaves
f. Infrared
g. Ultraviolet
h. X-Rays
i. gamma rays
2) A/C Transformer
The input voltage to a transformer is 120V RMS AC to the primary coil of 1000 turns. What are the
number of turns in the secondary needed to produce an output voltage of 10V RMS AC?
Given:
Input voltage of primary coil = Vp = 120V
# of turns in primary coil = Np = 1000 turns
Output voltage of secondary coil = Vs = 10V
Find: Number of turns in the secondary coil = Ns = ?
Formula:
Vp
Np
=
Vs
Ns
Ns =
Vs
× Np
Vp
Ns =
10
× 1000
120
Ns =
10
× 1000
120
𝐍𝐬 = 𝟖𝟑. 𝟑𝟑 𝒐𝒓 𝟖𝟑
Therefore, the number of turns in the secondary coil = Ns = 83 turns
Page 3
3) “Charging” the magnetic field of an inductor
60.000m of wire is wound on a cylinder, tight packed and without any overlap, to a diameter of 2.00 cm
(rsolenoid=0.0100m). The wire has a radius of rwire = 0.00100m and a total resistance of 0.325 . This
inductor initially has no current flowing in it. It is suddenly connected to a DC voltage source at time
t=0.000 sec. Vs=2.00Volts. After 2 time constants, the current across the inductor will be….
Hint: first find the inductor currents It=o, It=∞, …
Given:
T = 0.000 sec
Vs = 2.00Volts
rwire = 0.00100m
R = 0.325
The formula for the time constant is:
T =
L
R
After the 2 time constants:
T =
L
×2
R
The growth of the current will be:
𝑅
I = ( 1 − 𝑒 −( 𝐿 )𝑡
Multiply the T:
𝑅
𝐿
I = ( 1 − 𝑒 −( 𝐿 )𝑡 ×(𝑅×2)
I = ( 1 − 𝑒 −2 )
Therefore,
𝐈 = 𝟎. 𝟖𝟔𝟓 𝐈𝐨
Calculate the I
I = 0.865
I = 0.865 ×
Vs
R
2
0.325
𝐈 = 𝟓. 𝟑𝟐 𝑨
Therefore, the electric current = I = 5.32 A
Page 4
4) D/C Transformer
The input voltage to a transformer is 120V DC to the primary coil of 1000 turns. What are the number
of turns in the secondary needed to produce an output voltage of 10V DC?
Given:
Input voltage of primary coil = Vp = 120V
# of turns in primary coil = Np = 1000 turns
Output voltage of secondary coil = Vs = 10V
Find: Number of turns in the secondary coil = Ns = ?
Formula:
Vp
Np
=
Vs
Ns
Ns =
Vs
× Np
Vp
Ns =
10
× 1000
120
Ns =
10
× 1000
120
𝐍𝐬 = 𝟖𝟑. 𝟑𝟑 𝒐𝒓 𝟖𝟑
Therefore, the number of turns in the secondary coil = Ns = 83 turns
5)
Sound level of fireworks
At a fireworks show, a mortar shell explodes 25 m above the ground, momentarily radiating 75 kW of power as
sound. The sound radiates from the explosion equally efficiently in all directions. You are on the ground,
directly below the explosion. Calculate the sound level produced by the explosion, at your location.
Given:
P = 75kW or 75x103 W
r = 25m
Io = 1x10-12 W/M2
Calculate the intensity of the sound using this formula:
I =
I =
P
4πr 2
75 × 103
4π(25)2
Therefore, the intensity of the sound is:
𝐈 = 𝟗. 𝟓𝟓
𝑾
𝒎𝟐
Page 5
Calculate the sound level using this formula:
L = 10 𝑙𝑜𝑔
L = 10 𝑙𝑜𝑔
𝐼
𝐼𝑜
9.55
1 × 10−12
L = 10 (log (9.55) + 12)
L = 10 (log (9.55) + 12)
𝐋 = 𝟏𝟐𝟗. 𝟖𝟎𝟎𝒅𝑩
Therefore, the sound level produced by the explosion in your location = 129.800 dB
6) Dr Examines Image of a patients tiny mole w/ magnifying lens
A doctor (Veterinarian) examines a mole that is 15.3cm away from a magnifying lens, as shown
below. The lens has a focal length of 19.7cm. What is its magnification? Hint: Where is the image of
the mole?
Given:
u = -15.3cm
F = 19.7 cm
Using the lens formula,
1 1 1
= −
𝑓 V u
Substitute the given,
1
1
1
= −
19.7 𝑉 −1.53
1
1
1
15.3 − 19.7
=
−
=
𝑣 1.97 1.53 1.97 × 1.53
𝑣=
19.7 × 15.3
−4.4
Page 6
Therefore, the image distance, V is,
𝒗 = 𝟔𝟖. 𝟓𝟎 𝒄𝒎
Calculate the magmification using the formula
𝑚=
𝑚=
𝑣
𝑢
−68.50
−15.3
𝒎 = 𝟒. 𝟒𝟕𝟕
Therefore, the magnification is 4.477
7) Calculating with Faradays law and magnetic flux
A flat circular coil of wire has a radius of 0.18 m and is made of 75 turns of wire. The
coil is lying flat on a level surface and is entirely within a uniform magnetic field with a
magnitude of 0.55 T, pointing straight into the paper. The magnetic field is then
completely removed over a time duration of 0.050 s. Calculate the average magnitude of
the induced EMF during this time duration.
Given:
r = 0.18m
time duration = 0.050s
n = 75
B = 0.55 T
0.55 T
8
0.1
m...
