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.
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.
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.
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.
Combine the two to get an expression for area in
terms of volume
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.
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.
Y components of 2 and 3 cancel, so all we have to do
is add the x components
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?
speed = distance / Actual time = distance / observer's time x γ
Question: Three equal charges are placed at the corners of an
equilateral triangle. What is the force on each?
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
Adding the forces from the two charges, that is
Coulomb's law, force of attraction/repulsion
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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