Questions and Answers

Unstoppable object & immovable object
What constitutes a shock hazard
120 versus 240 volts
Speed of light and headlights
Minimize surface area of a cylinder
4 Spheres, gravitational force
Maximize volume of cylinder
Gravitational balance point
Weight versus altitude

Football kicked to goal
Rock dropped in well
energy needed to heat ice to steam
Four masses on the corners of a square
Relavistic traveler
Three charges in a triangle
Four charges in a square
2 charges in a line
Centrifugal force on earth
Chemical reactions

Question: What happens when an Unstoppable object meets an immovable object

Ans: This gets asked several times a week and gets boring.

Unstoppable objects and immovable objects cannot exist in the same universe, as each one assumes the other is not possible. In other words, an unstoppable object cannot exist in a universe that contains an immovable object and VV.

So the question is one of the many meaningless questions, such as those involving traveling faster than light, or temperatures lower that 0K, and many others.

Question: What constitutes a shock hazard

Ans: Here is one standard. Most are similar to this.
SHOCK HAZARD: As defined in American National Standard, C39.5, Safety Requirements for Electrical & Electronic Measuring & Controlling Instrumentation: A shock hazard shall be considered to exist at any part involving a potential in excess of 30 volts RMS (sine wave) or 42.4 volts DC or peak and where a leakage current from that part to ground exceeds 0.5 milliampere, when measured with an appropriate measuring instrument defined in Section 11.6.1 of ANSI C39.5.

Question: Why does the US use 120 volts while most of the world uses 240?

Ans: The answer is in the history. and politics.

The US was the first country to develop power distribution systems, and the first ones were built by Edison and his company. He built distribution systems of DC to light bulbs, and he picked 100 volts as the best compromise between difficulty in manufacturing and losses in distribution. Then he added 10-15 volts to the 100 for distribution losses to get 110 to 115 volts.

Higher voltage bulbs were more difficult to make, as you needed much thinner filaments.

AC power distribution beat out DC power distribution for many reasons, and they built on the Edison system, keeping the 115 volt level, but using transformers to make the distribution a lot more practical. There are lots of stories about AC (Westinghouse and Tesla) versus DC (Edison), commonly called "The war of currents", which Westinghouse won.

Then Europe started electrifying, and the equipment manufacturers in Europe wanted to use s different voltage so they could get a lock on supplying equipment without competition from US companies. They picked 230-240 volts at 50 Hz, twice the US as their voltage, and a slightly different frequency, reasoning that a higher voltage keep the size of wires in the house down, making it cheaper to wire the houses. Rest of world followed Europe.

Safety was never the issue, just industrial politics.

Question: If you are moving at the speed of light and turn on your headlights..

Ans: This gets asked several times a week.

It's impossible to move at the speed of light so the question is meaningless.

If you were moving at 99% of the speed of light, the light from your headlights would recede from you at the speed of light, or that is what you would perceive. That is what special relativity is all about. Velocities are always relative, and light always travels at the speed of light.

Question: Minimize surface area of a cylinder. Volume = V

Ans: cylinder V = πr²h
A = 2πr² + 2πrh

V = πr²h = V
h = (V/πr²)
A = 2πr² + 2πrh
A = 2πr² + 2πr(V/πr²)
A = 2πr² + 2V/r

to get a min/max, differentiate and set equal to 0
A' = 4πr – 2V/r² = 0
4πr³ – 2V = 0
r = ∛[ 2V/4π ] = ∛[V/2π]
d = 2r = ∛[4V/π]
h = V/πr² = ∛[V/4π]
h = 2r

Question: four identical spheres on the corners of a square, what is the gravitational force on one of the spheres due to the others

Ans: Side of square = s, mass of a sphere = m

The direction is directly towards the one diagonally opposite. You can easily calculate the force between the two on the diagonals, and then from one of the ones on the side and add all 3 via vector arithmetic.

For the one on the diagonal, the distance is s√2

Gravitational attraction in newtons
F = G m₁m₂/r²
G = 6.67e-11 m³/kgs²
m₁ and m₂ are the masses of the two objects in kg
r is the distance in meters between their centers

F₁ = Gm²/(s√2)² = Gm²/2s²

for one of the closer ones
F = Gm²/s²
but only the component parallel to the diagonal counts
The component is the force multiplied by cos45º or 1/√2
so that is
F₂ = 2•(1/√2)Gm²/s² = (√2)Gm²/s²

Total force is the sum F₁ + 2F₂
Ft = Gm²/2s² + 2(√2)Gm²/s²
Ft = [Gm²/s²][(1/2)+2√2]
Ft = [Gm²/s²](3.33) = (m/s)²(2.22e-10)

Question: What are proportions of a cylinder that will maximize volume for a given surface area

Ans: cylinder V = πr²h
A = 2πr² + 2πrh = 2πr(r+h)

Combine the two to get an expression for area in terms of volume
A = 2πr² + 2πrh
h = V/πr²
A = 2πr² + 2πr(V/πr²)
A = 2πr² + (2V/r)
if you differentiate and set equal to zero, you can find
A' = 4πr – 2V/r² = 0
4πr³ – 2V = 0
4πr³ = 2V
r = ∛[ V/2π ]
h = V/πr² = V/[π(V/2π)^(2/3) ]
h = (V/π)(2π/V)^(2/3)
h = (V/π)∛(4π²/V²)
the ratio is
h/r = (V/π)∛(4π²/V²) / ∛[ V/2π ]
h/r = (V/π)∛(4π²/V²) * ∛[ 2π/V ]
h/r = (V/π)∛(2³π³/V³)
h/r = (V/π)(2π/V) = 2

since diameter is 2r, this can also be stated as
h/d = 1


Ans: Lasers are very dangerous. Specially pointing it at the sky, as it if hits an aircraft, you are in big trouble. The department of homeland security has classified that as a terrorist act. There are hundreds of incidents where a pilot was partially blinded and the co-pilot had to take over. If that occurs at a time where the co-pilot were not available, you can have a crash and kill hundreds.

In general powerful lasers can be used only indoors, in an windowless room with controlled access, and anyone in the room must use the correct safety glasses (which vary depending on type of laser).

The problem is not just the beam hitting someone's eye, the problem is that the beam can be reflected off any shiny object, like a piece of metal or glass, travel through a window, and still damage someone's eye.

Lasers are NOT toys. They are dangerous instruments and must be handled correctly and with care.

Question: Find gravitational balance point between earth and moon.

Ans: Let x be distance in meters from center of earth to the equilibrium point.

Gravitational attraction
F = G m₁m₂/r²
G = 6.67e-11 m³/kgs²
m₁ and m₂ are the masses of the two objects in kg
r is the distance in meters between their centers
(center of mass)
earth radius 6,371,000 m
earth mass 5.97e24 kg
moon mass 7.35e22 kg
moon distance 3.91e8 m

take a hypothetical 1 kg mass between the earth and moon. Set the force of gravity between the object and earth and the object and the moon equal. x is distance between test mass and earth center.

F1 = G Me/x² (attraction to earth, mass of earth is Me)
F2 = G Mm/(d-x)² (attraction to moon, mass of moon is Mm,
d is distance earth to moon, center to center)

set them equal
Me/x² = Mm/(d-x)²
putting in numbers and solve for x
Me/Mm = x²/(3.91e8-x)² = 81.2
1241e16 + 81.2x² – 635e8x – x² = 0
80.2x² – 635e8x + 1241e16 = 0
x² – 7.92e8x + 15.47e16 = 0
using an online quadratic solver
x = 4.42e8, 3.50e8 meters
the first is not between earth and moon, so answer is
x = 3.50e8 meters

What is the acceleration due to Earth's gravity at that point
a = Gm/r²
a is acceleration from mass attraction in m/s²
m is mass of earth or other body generating the a
G = 6.67e-11 m³/kgs²
earth GM = 3.987e14

a = (3.987e14) / (3.50e8)²
a = 0.00325 m/s²

What is the acceleration due to moon's gravity at that point
a = Gm/r²
a is acceleration from mass attraction in m/s²
m is mass of earth or other body generating the a
G = 6.67e-11 m³/kgs²
moon mass 7.35e22 kg

a = (6.67e-11)(7.35e22) / (3.91e8–3.50e8)²
a = 4.90e12) / (0.41e8)²
a = 0.0029 m/s² (error is due to the subtraction of two similar numbers)

Question: At what altitude above the surface of the Earth is your weight one-half your weight on the surface? Express your answer as a multiple of Earth's radius r.

Ans: Gravitational attraction in newtons
F = G m₁m₂/r²
G = 6.67e-11 m³/kgs²
m₁ and m₂ are the masses of the two objects in kg
r is the distance in meters between their centers
earth radius 6,371 km = 6.37e6 meters
earth mass 5.97e24 kg

F on surface = 2F at altitude
G m₁m₂/r² = 2G m₁m₂/ (r+h)²
1/r² = 2/(r+h)²
(r+h)² = 2r²
r² + 2rh + h² = 2r²
h² + 2rh – r² = 0
solve for h

quadratic equation:
to solve ax² + bx + c = 0
x = [–b ±√(b²–4ac)] / 2a
h = [–2r ±√(4r²+4r²)] / 2
h = [–r ± r√2]

skipping the neg answer
h = 0.414r

Question: Ball is kicked 25m from goalpost at 22 m/s angle 46º. Will it pass over a 2.9m goal?

Ans: Vx = 22 cos 46 = 15.3 m/s
Vy = 22 sin 46 = 15.8 m/s
time to reach the goal is 25m / 15.3 m/s = 1.64 seconds

In that time, the ball will go up passing the goal on the way, OR go up, then on the way back down pass the goal, or never pass the goal.

From what height will a dropped object hit a speed of 15.8 m/s?
v² = 2gh = 2•9.8•h
h = 12.73 meters
and how long will it take?
t = √(2h/g) = 1.61 seconds
This means the ball will rise to 12.73 meters, in 1.61 seconds

From the above, comparing to the 1.64s above, the ball is just past it's peak of 12.73 m. So, how far will it fall in the (1.64–1.61= 0.03s) remaining? h = ½gt² = ½9.8•0.03² = 0.004 meters

So the ball will pass the goal at 12.73 – 0.004m = 12.73 meters, above the goal by 12.7 – 2.9 = 9.8 meters

Question: Object is dropped into a well and the sound is heard 3 seconds later. How deep is the well? Use 340 m/s for speed of sound

Ans: The 3 seconds is the sum of two times:
1. time for the rock to fall, and that is d = ½gt² or t = √(2d/g)
2. time for the sound to return up the cliff, and that is t = d/340
The total time is
√(2d/9.8) + d/340 = 3
solve for d
√(2d/9.8) = 3 – d/340
square both sides
d/4.9 = 9 + d²/340² – 6d/340
multiply by 340²
23592d = 9•340² + d² –6d•340
d² – 25632d + 1040400 = 0
solve via quadratic equation
d = 40.65 meters

Question: What is the energy needed to heat 1 kg of ice at –20ºC to steam at 150ºC

specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
specific heat of steam is 2.1 kJ/kgK
heat of fusion of ice is 334 kJ/kg
heat of vaporization of water is 2256 kJ/kg

5 parts to this problem, add them up to get the total

to warm ice to 0ºC
E₁ = 2.06 kJ/kgC x 1 kg x 20ºC
to melt ice
E₂ = 334 kJ/kg x 1 kg
to warm water from 0ºC to 100ºC
E₃ = 4.186 kJ/kgC x 1 kg x 100ºC
to boil water
E₄ = 2256 kJ/kg x 1 kg
to heat steam from 100ºC to 150ºC
E₅ = 2.1 kJ/kgC x 1 kg x (150–100)ºC

E = E₁ + E₂ + E₃ + E₄ + E₅

Question: Four 1 kg masses on the corners of a 1 meter square what is the force on one of these?

Ans: Calculate them one at a time, and use vector arithmetic to add. Put the test mass at the origin, and the diagonal of the square along the +x axis.

Gravitational attraction in newtons
F = G m₁m₂/r²
G = 6.674e-11 m³/kgs²
m₁ and m₂ are the masses of the two objects in kg r is the distance in meters between their centers

#1, at +x = 1√2 meter distance
F1 = G(1)² / (√2)² = G/2
angle is 0º

#2 at distance 1 meter
F2 = G(1)² / (1)² = G
angle is 45º

#3 at distance 1 meter
same as F2 but angle is –45º

Y components of 2 and 3 cancel, so all we have to do is add the x components
F = F1 + F2cos45 + F3cos45
F = F1 + 1.414 F2
F = G/2 + 1.414G = 1.914G = 1.277e-10 Newtons

Question: At what constant velocity must a spacecraft travel from Earth if it is to reach a star 4.3 LY away in 3.8 years, as measured by travelers on the spacecraft?

Ans: Actual time = observer's time x γ, where γ is the relativistic time contraction factor, γ = 1/√(1–V²/c²)

speed = distance / Actual time = distance / observer's time x γ
speed = V = 4.3 LY / (3.8Y x γ)
but since we are using distance in LY and time in Y, speed is LY/Y, or actually V/c

V/c = 1.1316 (1/γ) = 1.1316√(1–V²/c²)
V²/c² = 1.2805(1–V²/c²)
V²/c² = 1.2805–(1.2805)(V²/c²)
(2.2805)V²/c² = 1.2805
V²/c² = 0.5615
V/c = 0.7493
or 75% of light speed or 2.246e8 m/s

γ = 1/√(1–V²/c²) = 1/√(1–0.5615) = 1.510
At that speed it takes (outside observer) 4.3/0.75 = 5.73 years to make the trip.
relativistic contraction changes that, for the traveler, to 5.73/1.51 = 3.80 years, check

Question: Three equal charges are placed at the corners of an equilateral triangle. What is the force on each?

Ans: The forces on each are equal from symmetry and are pointing directly away from the midpoint of the opposite side. D is the length of a side of the triangle.
The Force from one charge to either of the others is
F = kQ²/D²

From symmetry, the force is on a line bisecting the opposite side. The components orthogonal to that line cancel, so the net force from each is
F = (1/√2)kQ²/D² = (√2/2)kQ²/D²

Adding the forces from the two charges, that is
F = (√2)kQ²/D²

Coulomb's law, force of attraction/repulsion
F = kQ₁Q₂/r²
Q₁ and Q₂ are the charges in coulombs
r is separation in meters
k = 8.99e9 Nm²/C²

Question: Four equal charges are placed at the corners of a square. all are positive, all have the same charge. what is the force on each charge?

Ans: From symmetry, the force on each charge is identical, just differing in direction, so we just have to solve for one of them. In each case, the direction points away from the opposite charge.

Set the x axis running diagonally from the test charge to the diagonally opposite charge, with the test charge at x=0.

Let charge = q, side of square = s
For the charge opposite, distance is s√2
F = kq²/(s√2)² = kq²/2s², pointing away from the other charge

For the other two, the x component of the force is all we need, the y components cancel out.
F = (kq²/s²)cos45º = (kq²/s²)√2 / 2
there are two so the total is √2(kq²/s²)
adding them up, we get (kq²/s²)(1/2 + √2)

Coulomb's law, force of attraction/repulsion
F = kQ₁Q₂/r²
Q₁ and Q₂ are the charges in coulombs
r is separation in meters
k = 8.99e9 Nm²/C²

Question: Two charges q and Q, same polarity. Where is field between them zero. Separation is K, put on x axis with q at origin.

field due to q, E1 = kq/x²
field due to Q, E2 = kQ/(K–x)²
they have to be equal
kq/x² = kQ/(K–x)²
q(K–x)² = Qx²
q(K²+x²–2Kx) = Qx²
qK² + qx² – 2qKx = Qx²
x²(q–Q) – 2qKx + qK² = 0

quadratic equation:
x = [–b ±√(b²–4ac)] / 2a
x = [2qK ±√(4q²K²–4qK²(q–Q))] / 2(q–Q)
x = K[q ± √(q²–q(q–Q))] / (q–Q)
x = K[q ± √(qQ))] / (q–Q)

Electric field E = kQ/r²
in Newtons/coulomb OR volts/meter
k = 8.99e9 Nm²/C²

Question: What is the centrifugal force on a weight on earth at θ degrees north latitude

At the equator, you can use
Centripetal force f = mV²/r = mrω²
ω is angular velocity in radians/sec
1 radian/sec = 9.55 rev/min
m is mass in kg
r is radius of circle in meters
V is the tangental velocity in m/s
f is in Newtons
with r the radius of the earth and ω the rotation rate of the earth, 1 rev/24 hours, converted into rad/s
earth radius 6,371 km = 6.37e6 meters

Off the equator, r becomes the distance from the surface of earth to the axis of rotation, less than the radius. A little trig will get you that
cos θ = r/R where r is distance to axis and R is radius of earth
So we have F = mω²Rcosθ

1 rev/24hour = 1 rev/86400 sec
1 rev = 2π rad
therefore ω = (1 rev/86400 sec) x (2π rad/rev) = 0.00007272 rad/s
F = (0.0337)mcosθ

Question: what are masses of reactants in these reactions

2C₄H₁₀ + 13O₂ ➜ 8CO₂ + 10H₂O
atomic mass
C = 12
H = 1
O = 16
2C₄H₁₀ = 2•58 = 116
13O₂ = 13•32 = 416
8CO₂ = 8•44 = 352
10H₂O = 10•18 = 180
check 116+416 = 352+180 = 532

116 grams of C₄H₁₀ + 416 grams of O₂ ➜ 352 grams of CO₂ + 180 grams of H₂O
2 mole of C₄H₁₀ + 13 moles of O₂ ➜ 8 mole of CO₂ + 10 moles of H₂O
note that I used rounded values for atomic mass. You can get a more accurate
value if you use more precision, such as 12.011 for C.

C₂H₄ + 3O₂ ➜ 2CO₂ + 2H₂O
atomic mass
C = 12
H = 1
O = 16
C₂H₄ = 2•12+4 = 28
3O₂ = 3•2•16 = 96
2CO₂ = 2•(12+32) = 88
2H₂O = 2•18 = 36
check 28+96 = 88+36 = 124

1 mole of C₂H₄ + 3 moles of O₂ ➜ 2 mole of CO₂ + 2 moles of H₂O
28 grams of C₂H₄ + 96 grams of O₂ ➜ 88 grams of CO₂ + 36 grams of H₂O


CO + 2H₂ ➜ CH₃OH
atomic mass
C = 12
H = 1
O = 16
CO = 28
2H₂ = 4
CH₃OH = 12+3+16+1 = 32
1 mole of CO + 2 moles of H₂ ➜ 1 mole of CH₃OH
28 grams of CO + 4 grams of H₂ ➜ 32 grams of CH₃OH


CH₄ + 2O₂ ➜ CO₂ + 2H₂O
atomic mass
C = 12
H = 1
O = 16
CH₄ = 12+4 = 16
2O₂ = 4•16 = 64
CO₂ = 12+32 = 44
2H₂O = 2•18 = 36
check 16+64 = 44+36 = 80

1 mole of CH₄ + 2 moles of O₂ ➜ 1 mole of CO₂ + 2 moles of H₂O
16 grams of CH₄ + 64 grams of O₂ ➜ 44 grams of CO₂ + 36 grams of H₂O

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