GEAS
Q: One rebar has a length of 20 m and a diameter of 25
mm. Another rebar has an unknown length with a
diameter of 20 mm. Find its length if the two bars have
eQual amount of heat expended.
Solution:
𝐿2
𝐷2 2
=( )
𝐿1
𝐷1
𝐿2
20 2
=( )
2
25
𝑳𝟐 = 𝟏. 𝟐𝟖 𝒎
Q: Convert 20 ergs to eV.
Solution:
𝑒𝑉 = 20 (
6.24𝑥1011 𝑒𝑉
)
1𝑒𝑟𝑔
Q: Find the gravitational acceleration at a height of 6400
km above the ground.
Solution:
𝑟𝑒 2
𝑔𝑜 = 𝑔 (
)
𝑟𝑒 + ℎ
2
6400𝑘𝑚
𝑔𝑜 = 9.81 (
)
6400𝑘𝑚 + 6400𝑘𝑚
𝒈𝒐 = 𝟐. 𝟒𝟓 𝒎/𝒔
Note: The radius of the earth is 6400 km (approx.)
Q: Determine the safest velocity of a car as it moves
through a horizontal unbanked curve of radius 5.1 m.
μ=0.91.
Solution:
𝑉 = √𝑟μg
𝒆𝑽 = 𝟏. 𝟐𝟒𝟑𝒙𝟏𝟎𝟏𝟑 𝒆𝑽
𝑉 = √5.1(0.91)(9.81)
Q: Rate of dissolution of water is 13.26. Find the amount
of x that will give a neutral solution.
Solution:
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑠𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑝𝐻 + 𝑝𝑂𝐻
𝑽 = 𝟔. 𝟕𝟒 𝒎/𝒔
𝑝𝐻 + 𝑝𝑂𝐻 = 2𝑥
13.26 = 2𝑥
𝒙 = 𝟔. 𝟑𝟑 → 𝒏𝒆𝒖𝒕𝒓𝒂𝒍
Q: What is the amount of heat removed by the
refrigerator for 100 g of water at 20⁰C to turn it into ice at
-10⁰C.
Solution:
Step 1. Water turns to ice from 20⁰C to 0⁰C.
Use c = 1 cal/g⁰C  water
𝑄1 = 𝑚𝑐∆𝑇
𝑄1 = 100𝑔 (1
𝑔
) (00 𝐶 − 200 𝐶)
𝑐𝑎𝑙 0 𝐶
Q: Calculate the osmotic pressure of a 0.200 M solution of
non-ionic solute @ 25˚C.
Solution:
πV = nRT
n
𝑉
π = RT
= (0.200)(8.314)(25+273)
π = 495.54 Pa
Q: A pendulum swings 3 inches above its lowest position.
Calculate the velocity of the pendulum @ its lowest
position.
Solution:
Note: the velocity of the pendulum @ its lowest
position is maximum because all its potential
energy is converted to kinetic energy.
𝑄1 = 2000 𝑐𝑎𝑙
Step 2. Change of phase (LiQuid to solid). Lf=80cal/g
𝑄2 = 𝑚𝐿𝑓
𝑄2 = 100𝑔 (80
𝑐𝑎𝑙
)
𝑔
𝑄2 = 8000𝑐𝑎𝑙
Step 3. Change of Temperature from 0⁰C to -10⁰C.
Use c = 0.5 cal/g⁰C  ice
𝑄3 = 𝑚𝑐∆𝑇
𝑐𝑎𝑙
𝑄3 = 100𝑔 (0.5
) (−100 𝐶 − 00 𝐶)
𝑔
𝑄3 = 500𝑐𝑎𝑙
Step 4. Calculate Total Q.
𝑄𝑇 = 𝑄1 + 𝑄2 + 𝑄3
P.E. lost = K.E. gain
1
mgh = 2 mv2
v = √2𝑔ℎ
3
12
= √2(32.3) ( )
v = 4 ft/s
Q: How many nuclei are there in a 1 kg of Aluminum?
Solution:
1 𝑚𝑜𝑙
1kg Al x 26.98154𝑔 x NA (avogrado’s number)
= 2.232x1025
𝑄𝑇 = 2000 + 8000 + 500
𝑸𝑻 = 𝟏𝟎. 𝟓 𝒌𝒄𝒂𝒍
To God Be the Glory
Page 1
GEAS
Q: 1.5x1022 molecules of oxygen @ 20˚C is placed in a
container with a maximum capacity of 1L. What is the
pressure of the oxygen inside the container?
Solution:
PV = nkT
P=
=
𝑛𝑘𝑇
𝑉
18𝑔
𝑙𝑖𝑡𝑒𝑟
𝒈
𝑥
= 𝟏𝟖𝟎
1𝐿𝑖𝑡𝑒𝑟 0.1 𝑚𝑜𝑙
𝒎𝒐𝒍
Q: Calculate the force needed to move the wheel barrow
at an inclination of 10 degrees, loaded with 300 lbs.
Neglect the weight of the wheel barrow and the friction
of the plane. y
A: 𝟓𝟐. 𝟎𝟗 𝒍𝒃𝒔
(1.5𝑥1022 )(1.3806𝑥10−23 )(20+273)
(1𝑥10−3 )
P = 60.68 kPa
Q: A solid sphere (M=12 kg, R=8 cm) rolls without slipping
across an inclined plane from a height of 2m to the
ground level. What is the translational velocity of the
sphere as it rolls the ground?
Solution:
10
𝑣 = √ 𝑔ℎ
7
10
𝑣 = √ (9.81)(2)
7
𝒗 = 𝟓. 𝟐𝟗
Solution:
𝐹 = 300 sin 10°
𝐹 = 𝟓𝟐. 𝟎𝟗 𝒍𝒃𝒔
Q: A straight track of 1600 meters in length is being
trailed by a runner. He runs towards east completing the
length then returned and stopped half-way, for which
takes him a total of five minutes. Calculate the average
velocity of the runner. y
A: 8 m/s
Solution:
𝑣𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
1600 𝑚+800 𝑚
5 𝑚𝑖𝑛𝑠 (
60 𝑠
)
1 𝑚𝑖𝑛
= 8 m/s
𝒎
𝒔
Q: A certain liQuid with density of 6.3 g/cm3 at 25oC is
placed in a barometer instead of mercury. What is the
eQuivalent height of the liQuid measured at 1
atmosphere?
Solution:
1 atm = 760 mm of Hg = 76 cm
Q: A car of 3400 lbs was driven by a 150 lb-man encircling
a rotunda of 2000 ft radius, at a constant velocity of 60
mph. Calculate the Centrifugal force acting on the car. y
A: 426.88 lbs
Solution:
𝐹=
DHg = 13.6 g/cm3
Dli𝐐uid hli𝐐uid = DHg hHg
(6.03)hli𝐐uid = (16.3)(76 cm)
𝐹=
𝑚𝑣 2
𝑟
3400 𝑙𝑏+150 𝑙𝑏
𝑚𝑖𝑙𝑒𝑠 5280 𝑓𝑡
1 ℎ𝑟
(
)(60
(
)(
))2
𝑓𝑡
32.2 2
𝑠
ℎ𝑟
1 𝑚𝑖𝑙𝑒
3600 𝑠
2000 𝑓𝑡
𝐹 = 426.88 lbs
𝐡𝐥𝐢𝐪𝐮𝐢𝐝 = 𝟏𝟕𝟐 𝐜𝐦
Q: A solution is prepared by dissolving 9.0g of sugar in
500g of water. What is the molar mass of sugar at 25oC
when the osmotic pressure is 2.46 atm?
Solution:
𝜋 = CRT
π = osmotic pressure
C = concentration
R = 8.31 L·kPa/mol·K
Q: Find the first freQuency overtone of the instrument of
40 cm in length at 15 degrees Celsius. y
A: 𝟖𝟓𝟎 𝑯𝒛
Solution:
𝒗 = 𝟑𝟑𝟏 + 𝟎. 𝟔 ( 𝟏𝟓) 𝒎/𝒔
𝒗 = 𝟑𝟒𝟎 m/s
𝒇=
101.325 𝑘𝑃𝑎
1 𝑎𝑡𝑚
𝐶=
(8.31)(25 + 273)
2.46 𝑥
𝒗
𝝀
𝟑𝟒𝟎
𝟎.𝟒𝟎
𝒇=
𝑯𝒛
𝒇 = 𝟖𝟓𝟎 𝑯𝒛
𝐶 = 0.1 mol/liter
9.0g of sugar in 500g of water is eQuivalent to 18g of
18𝑔 𝑠𝑢𝑔𝑎𝑟
sugar in 1000g of water. Thus, 1000𝑔 𝑤𝑎𝑡𝑒𝑟 is eQual to
18𝑔 𝑠𝑢𝑔𝑎𝑟
.
1𝐿𝑖𝑡𝑒𝑟 𝑤𝑎𝑡𝑒𝑟
To God Be the Glory
Page 2
GEAS
Q. An automobile moves in a constant speed of 50
miles/hr around a 1 mile diameter track. Find the angular
speed and period.
A. 𝝎 = 𝟏𝟎𝟎 𝒓𝒂𝒅/𝒉𝒓 , T= 3.8 min
𝑚𝑖𝑙𝑒𝑠
𝑣 50 ℎ𝑟
𝜔= =
= 100 𝑟𝑎𝑑/ℎ𝑟
𝑟 . 5 𝑚𝑖𝑙𝑒𝑠
𝑝𝑒𝑟𝑖𝑜𝑑 =
2𝜋
2𝜋
=
= 0.0628 ℎ𝑟 = 3.8 𝑚𝑖𝑛
𝜔
100𝑟𝑎𝑑/ℎ𝑟
Q. A 80 cm long flexible wire has a mass of 0.40 g. It is
stretched to 50 cm apart by a force of 500N. Find the
freQuency with which the wire vibrates.
A. 1000 Hz
Solution:
T
500 N
v = √ = √0.4 x10−3 kg
μ
Solution:
PV = nRT
𝑃 𝑉
𝑇=
𝑅 𝑛
1
𝑎𝑡𝑚
𝑃: (10−16 ) (
) = 1.316𝑥10−19
760 𝑡𝑜𝑟𝑟
𝐿 . 𝑎𝑡𝑚
𝑅 = 0.821
𝑚𝑜𝑙 . 𝐾
3
(1000𝑐𝑚 ⁄𝐿)
4 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑛= (
)(
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
1 𝑐𝑚3
6.02𝑥1023
)
𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒𝑠
𝑛 = 6.642𝑥10−21 𝐿
(1.31𝑥10−19 )(1)
𝑇=(
= 241°𝐾 = 𝟑𝟏°𝑪
(0.0821)(6.642𝑥10−21 )
Q. A paratrooper jumps in the airplane from a certain
height above the ground. What is his terminal velocity?
His mass is 150kg and a drag force is k=0.5 N-s/m.
⁄0.8m
v = 1264
𝜆
L = 2 = 0.5 𝑚
λ = 2L = 1 m
f = v/λ = 1000/ 1 = 1000Hz
A. Vt =
Vt =
Q. A punching bag of mass 150 kg is attached to an
overhead beam by a rope 0.5 m long. Manny PacQuiao
hits the bag with a large force causing it to transmit pulse
to the beam. Find the velocity of the transverse pulse if
the rope weighs 500g.
A. 38 36 m/s
Solution:
𝒎𝒈
𝒌
150(9.81)
0.5
Vt = 2,943 m/s
Q. what is the molarity of ethanol content in a liter of a 90
proof whisky?
MW of ethanol is 46 g/mol, density of alcohol is 0.8 g/ml
A. From the definition, 100% alcohol is 200 proof,
𝒗=
𝑴𝒈𝑳
√
𝒎
𝒎
𝒔
(𝟏𝟓𝟎 𝒌𝒈)(𝟗.𝟖𝟏 𝟐)(𝟎.𝟓 𝒎)
=√
So , 90 proof is 45% alcohol
(.𝟓𝟎𝟎𝒈)
45% of 1 liter is 450ml.
V= 38 .36 m/s
Q. The radioactive decay constant of radium is
1.36𝑥10−11 . Calculate the disintegrations per second
that occurs in 100 g of radium.
No of moles =
100 𝑔
226
𝑔
𝑚𝑜𝑙𝑒
= 0.4425 𝑚𝑜𝑙𝑒𝑠
= 0.4425 𝑚𝑜𝑙𝑒𝑠 (6.02𝑥1023
= 2.665𝑥1023
𝑎𝑡𝑜𝑚𝑠
)
𝑚𝑜𝑙𝑒𝑠
𝑑 = 𝜆𝑅
𝑑 = (2.665𝑥1023 )(1.36𝑥10−11 )
d = 3.624𝑥1012 𝑑𝑖𝑠/ 𝑠𝑒𝑐.
Q. The pressure of interplanetary space is estimated to be
in the order of 10−16 𝑡𝑜𝑟𝑟. This corresponds to a density
of 4 molecules per cubic cm. Determine the temperature
assigned to the molecules for the interplanetary space.
Molarity =
(450𝑚𝑙)(
0.8𝑔 1
)( 𝑚𝑜𝑙/𝑔)
𝑚𝑙 46
1 𝑙𝑖𝑡𝑒𝑟
Molarity = 7.83 mol/liter
Q. Given a surface tension af a certain liQuid, Ý=76.76
dyne/cm, find the height of a capillary tube with a radius
of the opening to be 0.01cm. and a density of 1g/cc.
A. g=981cm/s 2
Ý=76.76 dyne/cm = 76.76g/s 2
p=density
Ý=
𝐫𝐩𝐠𝐡
𝟐
76.76 g/s 2 =
1g
cc
(0.01cm)( )(
981cm
)h
s2
2
h=15.65cm
A. 31℃
To God Be the Glory
Page 3
GEAS
Q. A certain wire has a radius of 0.2m is suspended and
acts as a pendulum. It has a mass of 0.5kg. find k in time
4s.
A. I =
I=
𝒎𝑹𝟐
𝟐
Soln:
(0.5𝑘𝑔)(0.2𝑚)2
2
𝟐
𝐸 𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝐸 𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑠𝑎𝑙𝑡
I = 0.01kg-𝒎
Find k.
𝑇 4 − (𝑇𝑎)4
(340)4 − (293)4
= 𝑇 4 − (𝑇𝑎)4 = (310)4 − (293)4
= 3.2
𝐈
t=4sec
Q: How much larger is the energy absorb of water at 340K
use as solar panel compare to salt 310K. assume the
ambient temperature is 293K
t= 2𝝅√𝐤
Q: what is the rest mass energy of electron in ergs?
K= 0.0247
Q. Aluminum of length 2m is suspended in the ceiling and
is allowed to carry a load of 5kg. It has a radius of 0.01m.
(By the thorough experiment made by Sir Mel Maceda in
Texas labs, it was found out that the modulus of elasticity
of aluminum is 6.9x1010 N/𝑚2 ). Find the length of
elongation caused by the load to the aluminum.
𝑭𝒍
A. 𝜟𝒍 = 𝑬𝑨
F = (5kg)(9.81m/𝑠 2 ) = 49N
A = 𝝅𝒓𝟐 = 𝜋(0.01)2
(49N)(2𝑚)
Solution:
E = mc2
E = (9.11 x 10-28 g)(3 x 1010 cm/s)2
E = 2.733 x 10-17 Erg
Ans.
Q: a 4 inch thick brick has one temperature 350 oF and
other temperature of 100 oF, what is the temperature
gradient of the brick?
Solution:
ΔT / ΔL = (T1 - T2) / L
ΔT / ΔL = (350 – 100) / 4
ΔT / ΔL = 62.5 oF / in Ans.
𝛥𝑙 = ( 6.9x1010 N/𝑚2 )(𝜋(0.01𝑚)2 )
Q: a large diamond from africa is 44.5 carrat. Find the
number of of atoms of carbon in that diamond.
𝜟𝒍 = . 𝟎𝟎𝟒𝟓𝒎𝒎
Q: Determine the ∆V of a sphere which has a radius of
1.5m. ∆P=3.0 x106 Pa and bulk modulus of 6.7 x1010 Pa.
Soln:
4
V= 3π(r3)
Solution:
1 carrat = 0.2 grams
carbon atomic mass = 12.011 grams / mol
mass = (44.5 carrat)(0.2 g / carrat)
= 8.9 g
4
= 3π(1.53) = 14.13 m3
∆𝑉 =
Vo ∆P
B
=
(14.13 )(3.0 x10^6)
6.7 x10^10
= 6.32 x 10-4 m
Q: Determine the wavelength of electron whose mass
“m” is moving “v”. Using deBroglie formula.
number of moles = (8.9 g) / (12.011 g / mol)
= 0.741 moles
number of atoms = (0.741 moles)(NA)
= 4.463 x 1023 atoms
Ans.
Soln:
𝜆=
ℎ
𝑚𝑣
ans
where h = planks constant
Q: Determine the wavelength of a 5gram object moving 1
m/s
Soln:
𝜆=
ℎ
𝑚𝑣
=
6.62 𝑥 10^−34
1𝑚
)
𝑠
(5𝑥103 )(
To God Be the Glory
= 1.324x10^-37m
Q: a watch repairman has a magnifying glass with focal
length 8 cm to view the part of watch which is 1.3 cm
wide. What is the maximum apparent size of the part of
the watch viewed on the lens?
Solution:
M = 1 + (25 cm / f)
M = 1 + (25 cm / 8 cm)
M = 4.1
Maximum apparent size = 4.1 x 1.3 cm
= 5.33 cm Ans.
Page 4
GEAS
Q: FORMULA TO USE: 𝐻 =
2𝜋𝐾𝐿(𝑇2 −𝑇1 )
𝑅
ln 2
𝑅1
K-thermal conductivity
L-length of the cylinder
R1-radius of the wire
R2-radius of the cylinder
T1-temperature in Kelvin of the wire
T2-temperature in Kelvin of the cylinder
How many calories are developed in 1min in an electric
heater which draws 5A that is connected on a 110V line?
(110)(5)(60)
𝐼𝑅 2 𝑡
𝑉𝐼𝑡
𝑄=
=
=
4.186 4.186
4.186
Q = 7883.421 cal
A 200lbs block is in contact on the horizontal plane and is
moved by a force of 100lbs. If its coefficient of kinetic
friction is 0.2. How 200lb long will it take to change the
velocity of 4ft/s to 10ft/s?
𝑤
∑ 𝐹𝑡 = (𝑣2 − 𝑣1 )
𝑔
(200 – 0.2(100))t = (100/32.2)(10 – 4)
t = 0.1035 seconds
The average kinetic energy of the molecular gas is
6.2x1021. Find the temperature of the gas molecule.
3
𝐸 = 𝐾𝑇
2
6.2x1021 = (3/2)(1.38x10-23)T
T = _______
Q: An amount is compounded continuously with an
annual interest rate of 6%. How many years is it
compounded if the compound amount factor is
5.57767282?
𝑒𝑚 − 1
= 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 𝑎𝑚𝑜𝑢𝑛𝑡 𝑓𝑎𝑐𝑡𝑜𝑟
𝑒𝑟 − 1
𝑒 𝑛(0.06) − 1
= 5.57767282
𝑒 0.06 − 1
A: n=5years
Q: A biker starts to lose control of his bike at an altitude
of 1480m. At this moment his velocity is 100m/s… Find his
velocity at 480m above sea level.
𝑉 2 = 𝑉0 2 + 2𝑔ℎ
𝑉 = √1002 + 2(9.8)(1480 − 480)
A: V = 172m/s
Q: How long will it take to heat up 100g of water by 50oC
using a 240V, 20-ohm heating element?
𝑄 = 𝑚𝑐∆𝑇
2880𝑊 = (100𝑔) (4.180
Q: What is the final value of P10000 after 10 years when
compounded continuously with a nominal interest at 3%.
𝐹 = 𝑃𝑒 𝑚
𝐹 = 10000𝑒 .03(10)
𝐽
𝑔0 𝐶
) (50𝑜 𝐶)
𝐽
𝐽
(2880 ) 𝑡 = (100𝑔) (4.180 0 ) (50𝑜 𝐶)
𝑠
𝑔 𝐶
A: t = 7.257s
Q: How much heat must be absorbed by a 250g ice at 80oC to be a liQuid at 22oC? Cice= 2.3 J/goC
A: P13498.60
Q1 = mcΔT = (250g)(2.3)(8) = 4600J
Q: What is 1amu in MeV
1𝑎𝑚𝑢 = 1𝑎𝑚𝑢 (
Q2 = mLf = (250g)(333.5 KJ/kg) = 83.375J
931𝑀𝑒𝑉
)
1𝑎𝑚𝑢
A: 1amu = 931MeV
Q: A specific ore is composed of 17.2% iron compound.
This compound is in turn composed of 69.9% iron. What
percent of the ore is iron?
{
17.2𝑔 𝑜𝑓 𝑖𝑟𝑜𝑛 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
69.9𝑔 𝑜𝑓 𝑖𝑟𝑜𝑛
(
)}
100𝑔 𝑜𝑓 𝑖𝑟𝑜𝑛 𝑜𝑟𝑒
100 𝑔 𝑜𝑓 𝑖𝑟𝑜𝑛 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
∗ 100% = %𝑖𝑟𝑜𝑛 𝑖𝑛 𝑜𝑟𝑒
Q3 = mcΔT = (250g)(4.184 J/goC)(22oC) = 23012J
A: QT = 22.7KJ
Q: What is the change in entropy of a system where… Tcold
= 52oC, Thot = 128oC… 2500J
∆𝑆 = ∆𝑆𝑐𝑜𝑙𝑑 − ∆𝑆ℎ𝑜𝑡
∆𝑆 =
2500𝐽
2500𝐽
−
(52 + 273)𝐾 (128 + 273)𝐾
A: ΔS = 1.46J/K
A: 12% of iron in ore
To God Be the Glory
Page 5
GEAS
Problem: What is the momentum of photon with
wavelength 20nm?
Solution:
ℎ 6.62𝑥10−34 𝐽 − 𝑠
𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = =
𝜆
20𝑥10−9 𝑚
= 𝟑. 𝟑𝒙𝟏𝟎−𝟐𝟔 𝒌𝒈 − 𝒎/𝒔
where:
ℎ = 𝑃𝐿𝑎𝑛𝑐𝑘 ′ 𝑠𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
ℎ = 6.62𝑥10−34 𝐽 − 𝑠
Problem: The solar constant or Quantity of radiation
received by earth from the sun is 0.14W/cm2. Assuming
that the sun may be regarded as ideal radiator, calculate
the surface temperature of the sun. The ratio of radius of
earth’s orbit to the radius of the sun is 216.
Solution:
Using Stefan’s Law,
𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
(𝑟⁄𝑅 )2
𝑇4 =
𝜎
𝑇4 =
0.14𝑊/𝑐𝑚2
5.6𝑥10−12
𝑊 𝐾
𝑐𝑚2 𝑑𝑒𝑔4
(216)2
𝑻 = 𝟓. 𝟖𝟒𝒙𝟏𝟎𝟑 𝑲
Problem: A 150 lb person runs up a 15 ft stairway in 10 s.
What is the hp rating of the person?
Solution:
𝑚𝑔ℎ (150)(15)
1ℎ𝑝
𝑓𝑡−𝑙𝑏
𝑝=
=
= 225 𝑠 𝑥
𝑡
10
550𝑓𝑡−𝑙𝑏
𝑠
= 𝟎. 𝟒𝟏𝒉𝒑
Problem: What is the maximum efficiency of steam
turbine where steam entering the turbine is heated to
655℃ and exhausted at 115℃ ?
Solution:
𝑇
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 1 − 𝑇 𝑐
Problem: Find the length of a meter stick (Lo = 1m) if it
travels at the speed of 3.7 meters per second.
Answer: 0.44m
Solution:
𝑐
𝑣
L = Lo √1 − ( )2
3𝑥10^8
L = Lo √1 − ( 3.7 𝑥 10^8)2 = 0.44
Problem: What is the wavelength of a violet line (n=6) in a
hydrogen line spectra using the Balmer’s eQuation.
Answer: 4.102x10-7
Solution:
1
1
1
= 𝑅 ( 2 − 2)
𝜆
2
𝑛
1
1
1
= (1.097𝑥107 ) ( 2 − 2 )
𝜆
2
6
𝜆 = 4.102𝑥10−7
Note: for red line: n=4
For blue – green: n=3
Problem: How much heat is reQuired to change a
temperature of 3 moles of mono-atomic ideal gas of 55K
at a constant pressure.
Answer: 3.427 kJ
Solution:
5
𝑄 = 𝑛𝑅𝑇
2
𝑄=
5
𝐽
(3 𝑚𝑜𝑙𝑒𝑠) (8.31
) (55)
2
𝑚𝑜𝑙. 𝐾
𝑄 = 3.427𝑘𝐽
Problem: What is the velocity of the Leo satellite with an
altitude of 15000km?
Answer: 4339.58m/s
𝐻
(115+273)
Solution:
2
𝑟𝑒
𝑔 = 𝑔𝑜 (
)
𝑟𝑒 + ℎ
2
6400𝑘𝑚
𝑔 = 9.81 (
)
6400𝑘𝑚 + 15000𝑘𝑚
𝑔 = 0.88 𝑚/𝑠 2
= 1 − (655+273) = 𝟎. 𝟓𝟖𝟐
Note: temperature must be in Kelvin
Problem: Find the velocity of sound (in m/s) at room
temperature.
Answer: 343 m/s
Solution:
.6
𝑉 = 331 +
𝑇
𝑑𝑒𝑔𝑟𝑒𝑒 𝐶𝑒𝑙𝑠𝑖𝑢𝑠
.6
𝑉 = 331 + 𝑑𝑒𝑔𝑟𝑒𝑒 𝐶𝑒𝑙𝑠𝑖𝑢𝑠 (20)= 343m/s
Problem: As 1 kg of air evaporates, it exerts 1.7 x 10^5 of
work on the atmosphere. What is the change in volume?
Answer: 1.7 m^3
Solution:
𝑊
1.7 𝑥 10^5
𝑉′ − 𝑉 =
=
= 1.7
𝑃
1.1 𝑥 10^5
To God Be the Glory
𝑣 = √𝑔𝑟
𝑣 = √0.88 ( 6400𝑘𝑚 + 15000𝑘𝑚)
𝑣 = 4339.58 𝑚/𝑠
Q: A pendulum 85cm long oscillates at a period of 1.6s.
What is the gravitational acceleration for this condition?
A: Sol’n:
T  2
L
g
0.85
g
m
Ans. g  13 2
s
1.6  2
Page 6
GEAS
Problem: A pendulum of 3ft long has one end attached to
a fixed point and the other end carries a 4lb block. When
then pendulum swings back and forth, it was found out
that the velocity is 10 ft/s, at a position which the string is
at an angle of 45. What is the tension of the string at that
instant?
Answer: 6.995 lb.ft
Solution:
𝑚𝑣 2
𝑇=
+ 𝑚𝑔𝑐𝑜𝑠𝜃
𝑅
2
4 10
𝑇=
(
) + 4 cos 45°
32 3
𝑇 = 6.995𝑙𝑏. 𝑓𝑡
Q: A stone is dropped into the well. Three seconds later,
the sound is heard from the top of the well. What is the
distance between the top of the well and the water
surface? Assume sound travels at 1100 ft/s
A: Sol’n:
Let t1 = time in going down
Q: The string of the conical pendulum is 10ft and the bob
has a mass of ½ slug. The pendulum is rotated ½ rev/s.
Find the angle of the string make with the vertical.
A: Sol’n:
cos  
cos  
Ans.
exhaust is cooled at 12 C ?
A: Sol’n:
Ans.
2y
EQuation 2
g
t1 
In going up,
t2 
y
Vsound

y
EQuation 3
1100
Substitute EQuation 2 and 3 to EQuation 1:
2y
y

3
g 1100
Eff  1 
TC
TH
Eff  1 
12  273
180  273
Eff  37.1%
Q: A curve in a road has an arc w/ 150ft radius. If the
roadbend is 30ft wide and the outer bound is 4ft higher
than the inner bound, for what velocity is the road ideally
banked?
A: Sol’n:
x 2 42  302
x  884
tan  
Solving y using Q.E. we get,
v  R  g  tan  
Ans.
y  132.6 ft
Ans. v  25
Q: A straight track is 1600m length. A runner begins at the
starting line, runs due east for full length, turns around
and run halfway back. The time for this run is 5min. What
is the average velocity of the runner?
A: Sol’n:
Vave
displacement

elapsedtime
1600  800
 60s 
5 min 

 1min 
m
Vave  2.7
s
Vave 
Ans.
To God Be the Glory
 0.324
superheated steam at a temperature of 180 C and it
EQuation 1
1 2
gt
2
2
Q: What is the efficiency of the turbine that operates in a
In going down, it is free-falling
y
 1 rev 
4 10  

2 s 
  71
2
t 2 = time in going up
t1  t2  3
mg
4 mL n 2
32
2
4
 0.135
884
150  32.2  0.135   25
ft
s
ft
s
Q: What thermal gradient is to be applied to aluminum
(2cmx4cm) if the heat flow is to be 100 J/s. The thermal
conductivity of aluminum is 205
J
s  m  deg
A: Sol’n:
Q
T
 t 
L kA 205
100
J
s
J
 0.02 x0.04 
s  m  deg
T
deg
 610
Ans.
L
m
 610
deg
m
Page 7
GEAS
Q: A piston and a cylinder are used to compress a 1 mol of
ideal gas from 0.010 m3 to 0.005 m3. If the system is kept
at 150 degree Celsius, how much is the heat released?
A: Sol’n:
 Vf 
Q  nRT ln  
 Vi 
J 

 0.005 
Q  1mol  8.314
 150  273 ln 

mol  K 

 0.010 
Ans. Q  2, 400 J
Q: 10 calories of heat was added to a 1 gram of water.
Calculate the change in mass of water.
A: 4.67 x 10-10
∆E = 10 calories = 4.19 x 107 ergs
∆E = ∆mc2
∆m =
∆E
c2
4.19 x 107
= (3 x 108 )2 = 4.67 x 10-10
Q: What is the speed of light passing through a glass
medium (n=1.5)?
A: 200 x 106
c
v=n=
3 x 108
1.5
= 200 x 106
Q: What is the force reQuired to produce C’ (512 Hz)
given that 100N is needed to produce C (256 Hz)?
A: 400N
𝐟 ′𝟐
= 𝟐
𝐟
f′2
F’= 2 F
f
Q: A solid circular disk with a weight of 32.2 lbs and a
diameter of 3ft. is suspended from a vertical wire on its
midpoint horizontally. The wire has a diameter of
5122
)
2562
= 100 (
= 400N
Q: A refrigerator extracts heat at 250 joules per cycle. If it
works at 53 joules per cycle, find the coefficient of
performance.
A. 4.7
Solution:
𝐐
𝐖=
𝐊
where: W – work
Q – heat extracted
K – coefficieny of performance
250J
K
250J
K =
= 4.7
53J
Q: Given the data above, compute the heat discharged in
a room.
A: 303 Joules
QDISCHARGED = 53 + 250 = 303 J
53J =
To God Be the Glory
inch
and 3ft long. It’s modulus of elasticity is G=12 lb/ft2.
Calculate the period of torsional vibration.
A: 339.29 x 103
Radius of disk 1.5 ft = 18 inches
L = 3ft = 36 inches
1
G = 12 lb/ft2 = 12 lb/inch2
Idisk =
Jwire =
𝐖 𝟐
𝐫
𝐠
𝟐
𝛑𝐃𝟒
𝟑𝟐
32.2
(18)2
= 32.2 2
=
1
8
π( )4
32
= 162
= 2.4 x 10-5
𝐈𝐋
𝐉𝐆
162(36)
T = 𝟐𝝅√ = 2𝜋√
(2.4 x 10−5)(
1
)
12
= 339.29 x 103
Q: A scientist is studying a specimen which has 1.3 cm
width and is using a magnifying lens with focal length of 8
cm. What is the largest apparent image that the lens can
produce?
A: 5.3625
For maximum magnification,
M=1+
𝐅′
𝐅
1
8
M=1+
𝟐𝟓 𝐜𝐦
𝐟
25 cm
=
8 cm
4.125
Image size = 1.3 x 4.125 = 5.3625
Q: A company must relocate one of its factories in three
years. The eQuipment for loading dock is considered for
purchase. The initial cost of the eQuipment is P20, 000
and its salvage value after three years is P8, 000. If the
company’s rate of return in money is 10%, calculate the
capital recovery per year.
A. P5, 625.38
Solution:
𝐂𝐚𝐩𝐢𝐭𝐚𝐥 𝐫𝐞𝐜𝐨𝐯𝐞𝐫𝐲 𝐩𝐞𝐫 𝐲𝐞𝐚𝐫
= 𝐚𝐧𝐧𝐮𝐚𝐥 𝐢𝐧𝐭𝐞𝐫𝐞𝐬𝐭
+ 𝐚𝐧𝐧𝐮𝐚𝐥 𝐝𝐞𝐩𝐫𝐞𝐜𝐢𝐚𝐭𝐢𝐨𝐧
Capital recovery per year
= (20, 000)(0.10)
(20, 000 − 8, 000)(0.10)
+
(1 + 0.10)3 − 1
= P5, 625.38
Page 8
GEAS
Q: If the wavelength of a stone moving at 1 m/s is
1.3x1031 m, find the mass of the stone.
A. 5g
Solution:
𝒉
𝝀=
𝒎𝒗
where: ℎ − 𝑝𝑙𝑎𝑛𝑘 ′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
6.626 × 10−34 J. s
1.3 × 1031 =
m
m (1 s )
m = 0.005 kg = 5g
Q: The sheets of brass and steel, each of thickness of 1cm,
are placed in contact. The outer surface of the brass is
kept at 100oC. The outer surface of the steel is kept at 0oC.
Find the temperature at the common interface. The
thermal conductivities of brass and steel are in the ratio
of 2:1.
A. 67oC
Solution:
𝐤𝟏
𝐓
=
𝐤 𝟐 𝟏𝟎𝟎°𝐂 − 𝐓
Where:
k1
=2
k2
T
2=
100°C − T
2(100° − T) = T
T = 66.67° = 67°
Q: Two moles of an ideal gas are compressed slowly and
isothermally from a volume of 4.0 to 1.0 ft3 at a
temperature of 300K. How much work is done?
A. 6900 kJ
Solution:
𝑾 = 𝟐. 𝟑𝟎𝟑𝒏𝑹𝑻𝟏 𝒍𝒐𝒈
W = 2.303(2) (8.314
𝑽𝟐
𝑽𝟏
pressure is 15
lb
in2
. How much work is done?
A. 4, 000 lb-ft
Solution:
Step 1: Solve the volume after expansion (boyle’s law)
lb
3
𝑷𝒂 𝑽𝒂 (15 + 14.7) (144 ft 2 ) (5 ft )
𝑽𝐛 =
=
lb
𝑷𝐛
(14.7) (144 2 )
ft
Vb = 10 ft3
Step 2: Solve the work done during isobaric
𝑽𝐛
𝑾𝒂𝐛 = 𝑷𝑽𝐛 𝒍𝒏 ( )
𝑽𝒂
lb
10
Wab = (14.7) (144 ft2 ) (10 ft 3 )ln ( 5 )
Wab =
15, 000 lb ∙ ft
Step 3: Solve the work done during volume change
𝑾𝐛𝐜 = 𝑷𝐛 (𝑽𝐛 − 𝑽𝒂 )
lb
Wbc = (14.7) (144 ft2 ) (10 ft − 5 ft) Wbc = 11, 000 lb ∙
ft
Step 4: Solve the net work
𝐖𝐚𝐜 = 𝐖𝐚𝐛 − 𝐖𝐛𝐜
Wac = 15, 000 − 11, 000
Wac = 4, 000 lb ∙ ft
Q: What is the change of entropy of 235 g of ice melts
into water?
A. 286 J/K
Since there is no given temperature, it is assumed that ice
melts at 0° C. the latent heat of fusion of the ice is 333
kJ/kg
kJ
1
) (300K)log ( )
mol∙K
4
W = 6900 kJ
Q: A rectangular drinking trough for animals is 2.1m long,
43.1cm wide, and 27cm deep. It contains 2.75kg with
specific volume of 1.7dm3/kg. What is the height?
ANS: 5.16 m
𝑽 = 𝒍𝒘𝒉
𝑉
ℎ=
𝑙𝑤
𝑑𝑚3
1𝑚3
(2.75𝑥105 ) (1.7
)(
)
𝑘𝑔
1000𝑑𝑚3
ℎ=
(2.1)(0.431)
𝒉 = 𝟓. 𝟏𝟔𝒎
Q: Air is expanded isothermally to atmospheric pressure
and then cooled y a constants pressure until it reaches its
initial volume. It occupies a volume of 5 ft3 and the gauge
𝑪𝒉𝒂𝒏𝒈𝒆 𝒐𝒇 𝒆𝒏𝒕𝒓𝒐𝒑𝒚 = 𝑸/𝑻 = 𝒎𝑳/𝑻
(0.235 𝑘𝑔)(333 𝑘𝐽/𝑘𝑔)
(0 + 273 𝐾)
𝑘𝐽
= 0.286
𝐾
= 𝟐𝟖𝟔 𝑱/𝑲
Q: A steam has an overall efficiency of 25%. What is the
rate at which heat is supplied to 2hp engine?
ANS: 20352 Βtu/hr
=
𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 =
𝑾
𝑸
𝑄=
𝑊
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦
𝑄=
2ℎ𝑝
0.25
2544𝛣𝑡𝑢/ℎ𝑟
)
1ℎ𝑝
= 8ℎ𝑝(
𝑸 = 𝟐𝟎𝟑𝟓𝟐 𝜝𝒕𝒖/𝒉𝒓
To God Be the Glory
Page 9
GEAS
Q: As explosives, TNT burns and releases energy of
4.2 𝑥 106 kJ/kg. The molecular weight of TNT is 227 kg.
Calculate the energy released by one molecule of TNT.
A. 9.87 eV
𝟏 𝒂𝒎𝒖 = 𝟏. 𝟔𝟔 𝒙 𝟏𝟎 −𝟐𝟕 𝒌𝒈
𝑬𝒏𝒆𝒓𝒈𝒚 𝒓𝒆𝒍𝒆𝒂𝒔𝒆𝒅
𝑘𝐽
(1.66 𝑥 10 −27 𝑥 227 𝑘𝑔)
𝑘𝑔
6.2415 𝑥 1018 𝑒𝑉
= 1.58 𝑥 10 –18 𝑘𝐽 (
)
1𝐽
= 𝟗. 𝟖𝟕 𝒆𝑽
Q: Calculate the wavelength in nanometer of x – ray
photon given the momentum of 6.62 𝑥 10 −26 𝑘𝑔 ∙ 𝑚/𝑠 2
.
A. 10 nm
𝒉
𝒑=
𝝀
𝑚
6.63 𝑥 10−34 𝐽 ∙ 𝑠
6.62 𝑥 10−26 𝑘𝑔 ∙ 2 =
𝑠
𝜆
–8
𝜆 = 1.001 𝑥 10 𝑚
𝝀 = 𝟏𝟎 𝒏𝒎
Q: Five kg of aluminum (𝐶𝑣 = 0.91 𝐽/𝑔𝑚 ∙ 𝐾) at 250 K is
placed in contact with 15 kg of copper (𝐶𝑣 = 0.36 𝐽/𝑔𝑚 ∙
𝐾) at 375 K. Assume there is no heat transfer, calculate
the final temperature.
A. 318 K
= 4.2 𝑥 106
𝑸𝑨𝒍 = 𝒎𝑪𝒗 ∆𝑻 = (5)(0.91)(𝑇 − 250 𝐾)
𝑸𝑪𝒖 = 𝒎𝑪𝒗 ∆𝑻 = (15)(0.36)(𝑇 − 375𝐾)
𝑸𝑨𝒍 + 𝑸𝑪𝒖 = 𝟎
(5)(0.91)(𝑇 − 250) + (15)(0.36)(𝑇 − 375) = 0
4.55 T – 1137.5 K + 5.4 T – 2025 K = 0
9.95 T = 3162.5 K
T = 318 K
Q: An FM broadcast at 98.1 MHz. If the transmitter has
5.0 𝑥 10 4 𝑊 power, calculate the emitted photons per
second.
A. 𝟕. 𝟔𝟖𝟕 𝒙 𝟏𝟎 𝟐𝟗 𝒑𝒉𝒐𝒕𝒐𝒏𝒔/𝒔
𝑬 = 𝒉𝒇
= (6.63 𝑥 10−34 𝐽 ∙ 𝑠)(98.1 𝑥 106 )
= 6.50403 𝑥 10−26 𝐽
Q: If the sun’s surface temperature increases by a factor
of 2, the resulting temperature of the earth’s atmosphere
would .
ANS: decrease by ¼
Let: T = temperature of the earth’s atmosphere
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒕𝒆𝒎𝒑. 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒖𝒏
𝟐𝒏
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑡𝑒𝑚𝑝. 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑛
𝑇=
22
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒕𝒆𝒎𝒑. 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒖𝒏
𝑻=
𝟒
𝑻=
Q: A 1kg body is converted entirely into energy. How long
will a light bulb be illuminated? A 100W light bulb is 100
energy.
ANS: 290322 years
𝑬 = 𝒎𝒄𝟐
𝐸 = (1𝑘𝑔)(3𝑥108 )2 = 9𝑥1016 J
Note: 1 light year = 3.1 x 107 s
1
9𝑥1012 𝑠
= 𝟐𝟗𝟎𝟑𝟐𝟐 𝒚𝒆𝒂𝒓𝒔
3.1𝑥107 𝑠/𝑦𝑒𝑎𝑟
Q: Calculate the volume of a 0.3N base to neutralize 3L of
0.01N nitric acid.
ANS: 0.1L
Note: Normality = eQuivalent /liter
For neutralization, the eQuivalent weight of the base is
eQual to the eQuivalent weight of the acid.
𝑵𝑨 𝑽 𝑨 = 𝑵𝜝 𝑽 𝜝
(0.01)(3) = (0.3)𝑉𝛣
𝑽𝜝 = 𝟎. 𝟏𝑳
Q: A 50kg 100 hp engine in a closed system is separated
from its surroundings. Find the time for the system to rise
to 10°C.
ANS: 28 sec
100hp x (746W/1hp) = 74600 W or J/s
1 cal = 4.186 J
𝐽
4
𝝆=
𝑷
5.0 𝑥 10 𝑊
=
𝑬
6.50403 𝑥 10−26 𝐽
𝝆 = 𝟕. 𝟔𝟖𝟕 𝒙 𝟏𝟎 𝟐𝟗 𝒑𝒉𝒐𝒕𝒐𝒏𝒔/𝒔
To God Be the Glory
1
9𝑥1016 𝐽 (100 𝐽/𝑠) (100 𝑒𝑛𝑒𝑟𝑔𝑦) = 9x1012s
1 𝑐𝑎𝑙
𝑃 = 74600 𝑠 𝑥 (4.186 𝐽) = 17821 cal/s
Rate (t) =mc∆T
(17821 cal/s) t = (5000g)(1 cal/g°C)(10°C)
t = 28s
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