One good sneeze in a quiet moment.
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The acceleration due to gravity on the earth, known as g, is 9.8m/s2 or 9.8m/s/s, which means that a falling object falls with an increasing velocity of 9.8m/s every second, or 9.8m/s/s, which means 9.8 meters/second/second. m/s/s can be written as m/s2, because m/s/s means m/s ÷ s/1 = m/s x 1/s = m/s2.
You can use Newton's equations of motion: s = ut + ½at² Ignoring air resistance: s = distance = height of cliff u = initial velocity = 0 m/s a = acceleration due to gravity ≈ 9.8 m/s² t = time = 5.86 s → height of cliff ≈ ½ × 9.8 m/s² × (5.86 s)² ≈ 168 m If you want it in feet: 0.3048 m = 1 ft (exactly) → 168 m = 168 m ÷ 0.3048 m/ft ≈ 551 ft
1 m/s² (1 metre per second per second) is a measure of acceleration, NOT speed. Assuming you mean a swim rate of 1.0 m/s (1 metre per second), then: time = distance ÷ speed = 900 m ÷ 1 m/s = 900 s 60 s = 1 min → 900 s = 900 ÷ 60 min = 15 minutes ---------------------------------- Assuming you really do mean that you swim with a constant acceleration of 1 m/s² and you start at 0 m/s, then: s = ut + ½at² s = 900 m u = 0 a = 1 m/s² t = unknown → 900 m = 0 m/s × t + ½ × 1 m/s² × t² → 900 m = ½t² → t² = 1800 s² → t ≈ 42.43 s If you start at 0 m/s and accelerate at 1 m/s², then it will take approx 42.43 seconds to cover 900 m.
1 g, or one gram of acceleration, is the acceleration experienced due to Earth's gravity, which is approximately 9.81 meters per second squared (m/s²). This means an object in free fall will increase its velocity by about 9.81 m/s for each second it falls. In terms of speed, this translates to a significant increase, as an object would reach about 39.2 m/s after 4 seconds of free fall, assuming no air resistance.
swim, gym, him, Guam
You do it the obvious way. Take p, q in Q, the rationals By definition, we can write p = m/n and q = r/s where m, r are integers, n and s are natural. we define pq (p times q) = (mr)/(ns) p/q = pq^-1 where q^-1 denotes q's multiplicative inverse s/r Remark: you cannot divide by 0 here because 1) 0 have no multiplicative inverse 2) if r = 0. s/r is undefined.
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Prove: [ P -> Q AND R -> S AND (P OR R) ] -> (Q OR S) -> NOT, --- 1. P -> Q ___ hypothesis 2. R -> S ___ hypothesis 3. P OR R ___ hypothesis 4. ~P OR Q ___ implication from hyp 1. 5. ~R OR S ___ implication from hyp 2 6. ~P OR Q OR S ___ addition to 4. 7. ~R OR Q OR S ___ addition to 5. 8. Let T == (Q OR S) ___ substitution 9. (~P OR T) AND (~R OR T) ___ Conjunction 6,7 10. T OR (~P AND ~R) ___ Distribution from 9 11. T OR ~(P OR R) ___ De Morgan's theorem 12. Let M == (P OR R) ___ substitution 13. (T OR ~M) AND M ___ conjunction 11, hyp 3 From there, you can use distribution to get (T AND M) OR (~M AND M). The contradiction goes away leaving you with T AND M, which can simplify to T.
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According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 1 words with the pattern M--Q--S-S. That is, nine letter words with 1st letter M and 4th letter Q and 7th letter S and 9th letter S. In alphabetical order, they are: marquises
B C D G J P R S U a b c d e f g j m n p q r s t and u * * * * * What? G, J, P, R, a, b, d, e, f, g, j ,m, n, p, q, r, t and u have no line of symmetry!
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 1 words with the pattern Q--GG-S. That is, seven letter words with 1st letter Q and 4th letter G and 5th letter G and 7th letter S. In alphabetical order, they are: quaggas
We can easily calculate that deceleration of block is a = v / t = 1.4 / 0.8 = 1.75 m/s2. The only force acting in the plane of movement may be Ft due to friction. Formula for Ft is Ft = Q * f, where Q is weight and f is a coefficient of friction. Q is: Q = m * g Ft has also to equal Ft = m * a, so: m * a = m * g * f a = g * f f = a / g f = v / ( t * g) For given data, f will be f = 1.4 / ( 0.8 * 9.81) = 0.178
"Squiggy" is spelled as S-Q-U-I-G-G-Y.
N = m*g N : newtons m : kilogramme g = 9.81 m/s²
G. M. Martinotti has written: 'Miogypsina in the Gaza 1 Well, Coastal Plain, Israel' -- subject(s): Miogypsina, Paleontology