1.A plastic strip has 7 black stripes and 7 clear stripes. The width of the blac
ID: 1899878 • Letter: 1
Question
1.A plastic strip has 7 black stripes and 7 clear stripes. The width of the black/clear pair is d = 4 cm (0.04 m). The strip is held just above the photogate and released. a) How much time does it take the first black/clear pair to fall through the photogate timer, starting from rest? b) What is the average velocity for the first pair (using the given d, and time from part a)? c) If the strip is held at an angle of 20 degrees from vertical, as shown in Figure 4, what is the vertical distance, y? d) How long does it take the strip to fall a distance y, starting from rest? e) If the computer calculates an average velocity by assuming that the strip was vertical and that the distance was d, not y, what velocity will it calculate for the first pair using the time in part (d)? f) What is the percentage error in velocity due to tilting the strip by 20 degrees? g) Will air resistance decrease or increase the measured acceleration? 2. a) How long would it take a cart dropped vertically to hit the floor if the bottom of the cart starts 1.5 m from the ground? (DonExplanation / Answer
a) starting from 0 m/s we accelerate due to gravity and travel a distance of 0.04m. We look throught our four kinematic equations to find one that gives time and only asks for starting speed, distance, and acceleration. d = vo*t + (1/2)gt^2 has what we want and doesn't need more than we can supply. t^2 = 2d/g t = SQRT(2d/g) = SQRT(2*0.04/9.8)s = 0.09s This makes sense: It takes a second to fall 4.5m, so we expect it to take a small fraction of a second to fall 4cm b)distance = average rate times time d = rt r = d/t = 0.04m/0.09s = 0.44m/s c) Now we fall a little longer distance. The long leg of the right triangle is 0.04m and the hypotenuse is 0.04m/cos(20) = 0.0426 m d) since we manioulated the equation first in part (a), we can substitiute our new value into that equation t = SQRT(2d/g) = SQRT(2*0.0426/9.8)s = 0.093s e) r = d/t = 0.04m/0.093s = 0.43m/s f) = (0.4427 - 0.429)/0.4427 = 0.03 = 3% g) air resistance slows down acceleration 2 use equation from 1(a) t = SQRT(2d/g) = SQRT(2*1.5/9.8)s = 0.55s (b) we are only getting 20cm/1.5m of the acceleration 9.8m/s^2*(.2/1.5) = 1.3 m/s^2 c)t = SQRT(2d/g) = SQRT(2*1.5/1.3)s = 1.5s d) friction will decrease acceleration
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.