求物理英文題目的解答?
2015-11-09 10:00 am
A 0.25 kg softball has a velocity of 15 m/s at an angle of 44° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a)24 m/s, vertically downward, and (b)24 m/s, horizontally back toward the pitcher?
回答 (1)
2015-11-09 12:19 pm
V1 = 15*(cos44*i - sin44*j)
= 15*(0.7193i - 0.6947j)
= 10.79i - 10.42j
(a) V2 = -24j
ΔP = P2 - P1
= 0.25*(-24j - 10.79i + 10.42j)
= -0.25*(10.79i + 13.58j)
= -(2.698i + 3.395j)
(b) V2 = -24i
ΔP = P2 - P1
= 0.25*(-24i - 10.79i + 10.42j)
= -8.698i + 2.605j
= 15*(0.7193i - 0.6947j)
= 10.79i - 10.42j
(a) V2 = -24j
ΔP = P2 - P1
= 0.25*(-24j - 10.79i + 10.42j)
= -0.25*(10.79i + 13.58j)
= -(2.698i + 3.395j)
(b) V2 = -24i
ΔP = P2 - P1
= 0.25*(-24i - 10.79i + 10.42j)
= -8.698i + 2.605j
收錄日期: 2021-04-30 20:04:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151109020013AAt5LeG