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bonufazy [111]
3 years ago
15

What is the Validity measures of a basketball bouncing

Physics
1 answer:
kykrilka [37]3 years ago
3 0

This study was aimed at testing the construct validity of the basketball basic motion skills test instrument (ITK GDBB). The research used descriptive method of 3 basketball experts in the city of Cimahi; 3 experts are the expert in basketball. The instrument used was the ITB GDBB developed by Silvy (2019) consisting of top passing, bottom passing, top service, bottom service, chest passing, bounding passing, overhead passing, and leading ball (dribbling). This instrument consists of 76 items that cover 4 domains in basketball, namely chest pass, overhead pass, bound pass, and dribbling. The validity method used the construct validity of different power types. For the reliability method, it used the Kuder Ricardson (KR) and Objectivity analysis. The results of the construct validity analysis of a total of 76 items show that the score is ranged from 0.67 to 1.00. The construct validity value of 71 items in the basketball game is in the high category (= 1.00), 5 items are in the sufficient category, the relativity score is ranged from 0.75 to 0.98, and the objectivity score is ranged from 0.89 to 0.95. The conclusion is that this test instrument can be used as a standardized basic motion skill test for standardized large ball games for validity in basic motion skills in basketball games for grade VII junior high school students.

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Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart  ⇒  X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds    ⇒   56lb

160pounds    ⇒ X

X=\frac{(160)(56)}{140} =64lb

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A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
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Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

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d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

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