show all work please 1. A plastic strip has 8 black stripes and 8 clear stripes.
ID: 1279263 • Letter: S
Question
show all work please
1. A plastic strip has 8 black stripes and 8 clear stripes. The width of the black/clear pair is d = 5 cm (0.05 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 30 degree 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)? What is the percentage error in velocity due to tilting the strip by 30 degree? 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 1.8 m from the ground? (Don't do this! the carts have delicate low friction bearings) b) If a track is 1.8 m long, and has a height change, Deltah, of 30 cm, what is the acceleration of a frictionless cart sliding down the track? c) How long will it take for a cart to travel the 1.8 m, starting from the top of the track? d) Will friction increase or decrease the acceleration?Explanation / Answer
a) d =gt^2/2 that is t = sqrt(2d/g) =sqrt(2*0.05/9.81) =0.100 sec
b) v = d/t =0.05/0.100=0.50 m/s
c) y = d*cos(30) =0.05*cos(30) =0.0433 m
d) y= g*t1^2/2 that is t1=sqrt(2y/g) =sqrt(2*0.0433/9.81)=0.0940 sec
e) v1 =d/t1 =0.05/0.0940=0.532 m/s
f) error = (v-v1)/v =(0.5-0.532)/0.5 =0.0643 =6.43%
g) Air will break the stick in its falling so that the total measured acceleration will be less than g=9.81 m/s^2
2)
a) h = gt^2/2 that is t = sqrt(2h/g) =sqrt(2*1.8/9.81) =0.606 sec
b) angle of slope with respect to horizontal is (L is slope length=1.8 m, H =0.3 m)
sin(alpha) = H/L
If there is no friction acceleration along the slope is
a = g*sin(alpha)=g*H/L =9.81*0.3/1.8 =1.635 m/s^2
c) L =a*t^2/2 that is t = sqrt(2*L/a) =sqrt(2*1.8/1.635) =1.484 sec
d) Friction force is opposing motion down the track therefore the total acceleration (downwards) will be reduced by friction.
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