f.3 physics 有兩題唔明!!(15點)

[匿名]

2010-08-15 1:11 am

Q11.
A parachutist is falling with a contast vertical velocity of 16 m/s but is being
horizontally at 12 m/s. What is his actual velocity?
If he jumps from a height of 500m, directly above where he intends to lands, how far away from this point does he actually land?

第一個o既question 我己經計左，但係第二個question 我唔係好明。

Q 9c) Table 9.4 shows the results as recorded by the student. However, on one occasion the student made an error in calculating the acceleration.

( i ) Plot a graph of cos θ (y-axis) against acceleration (x-axis).

( ii )Use the graph to determine
(1) which acceleration was calculated incorrectly,
(2) the correct value of this acceleration,
(3) the acceleration of the trolley when the elastic band is parallel to the
longer side of the trolley.

table9.4
--------------------------------------------------------------------------------------------------
Angle θ/----------------------------20------40------50------60-----70-----80

cos θ-------------------------------0.94---0.77---0.64---0.50---0.34---0.17

acceleration / cm/s²--------------61-----50------42------24------22-----11
--------------------------------------------------------------------------------------------------

我應該要點畫呢？

更新1:

Question 9c 用文字解釋都得

更新2:

I'm so sorry about the question(3), I' ll do it by myself. Thank you!!

回答 (1)

Godfrey

2010-08-16 6:23 pm

✔ 最佳答案

Q11.(a) The actual velocity is 20 m/s making angle 36.9degree with the vertical.(b) The direction of the actual velocity is 36.9degree with the vertical;The vertical displacement is 500 m,you have to find the horizontal displacement. This is the distance away from the intended point of landing.12/16 = x/500 m Where x is thedistance (off track)x = 500 (12/16) = 375 mAlternatively tan 36.9 degree =x/500It is 375 m from the original point wherehe is supposed to land.
Q 9c) (1) There are 6 points on the graph. We find the slope baseon two points, point of origin (0, 0) and other points in the table. It is astraight line passing (0, 0)0.94/64 = 0.0154, 0.77/50 = 0.0154, 0.64/42 = 0.0152,0.50/24=0.0208, 0.34/22 = 0.0154, 0.17/11 =0.0154The slope is close to 0.0153, with the exception of 0.50/24 = 0.0208. This point is off. 24 cm/s² was calculated incorrectly
(2). The graph can be represented by y = 0.0153 x The slope isestimated to be 0.0153 For the point cosθ = 0.5 0.5 = 0.0153 x x = 33 cm/s²

(3) Cannot do it without knowing procedures and the actual diagram for the experiment.

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