Answer:
2.4583 ± 0.0207 seconds
Explanation:
The time period of a pendulum is approximately given by the formula ...
T = 2π√(L/g)
The maximum period will be achieved when length is longest and gravity is smallest:
Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds
The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:
Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds
If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.
T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2
T ≈ 2.4583 ± 0.0207 . . . seconds
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We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.
Answer:
because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits
Answer:
horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s
initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s
Using s = ut + at2/2 for change in vertical distance in time t, with acceleration a (-9.8m/s2) and initial velocity u (vv = 3m/s) we have
0 = 3*t - 9.8*t2/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).
The horizontal distance in time t is vh*t or 3√3*6/9.8 m
Explanation:
Answer:In Example A, white light (i.e., a mixture of red, green and blue) shines upon a magenta filter. Magenta absorbs its complementary color - green. Thus, green is subtracted from white light.
A)♣ the string being pulled,
<span>the angular speed is: w1=w0 +w’*t, hence t=(w1-w0)/w’; </span>
<span>the angular path is: b=w0*t+0.5*w’*t^2, where angular acceleration </span>
<span>w’=T/J, torq T=F*r, and b*r=L; </span>
<span>♦ thus b=w0*(w1-w0)/w’ +0.5*w’*((w1-w0)/w’)^2 = </span>
<span>= (w1-w0)*(w0 +0.5w1 -0.5w0)/w’ =0.5*(w1^2 –w0^2)/w’; </span>
<span>♠ and b=L/r =0.5*(w1^2 –w0^2)/(F*r/J); </span>
<span>2(F*r/J)*L/r =w1^2 –w0^2, hence </span>
<span>w1^2=2F*L/J +w0^2; </span>
<span>b)♣ the power is P=F*(w0*r);
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