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ID: 1656801 • Letter: O
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ooo Verizon LTE 3:41 PM 43% loncapa.selu.edu You are correct Your receipt no. is 155-901 Previous Tries A shot putter accelerates the 7 kg shot in a straight line at an angle of 35° above the horizontal (the line of motion). The shot is accelerated from rest to a speed of 14 m/s in a time of 0.25 s What is the magnitude of the acceleration of the shot? 56.0 m/sA2 You are correct Previous Tries Your receipt no. is 155-1717 What is the magnitude of the component of the shot's weight parallel to the line of motion? 39.3 N You are correct Previous Tries Your receipt no. is 155-379 What is the magnitude of the component of the shot's weight perpendicular to the line of motion? 56.2 N You are correct Previous Tries Your receipt no. is 155-18696) What is the magnitude of the component of the shot putter's force parallel to the line of motion? ON Submit Answer Incorrect. Tries 4/99 Previous Tries What is the magnitude of the component of the shot putter's force perpendicular to the line of motion? ON Submit AnswerIncorrect. Tries 2/99 Previous Tries What is the angle of the shot putter's force relative to the line of motion? o Submit Answer Incorrect. Tries 1/99 Previous Tries What is the angle of the shot putter's force relativ the ntal? 125 Submit AnswerIncorrect. Tries 3/99 Previous Tries Threaded View Chronological View Sorting/Filtering options Export? NEW Preferences on what is marked as NEW Mark NEW posts no longer new Spencer Lee Shaw (W0413731:selu) Reply (Sat Sep 16 09:20:20 am 2017 (CDT)Explanation / Answer
parallel to line:
Fnet = F - m g sin35 =m a
F = 7 (9.8sin35 + 56)
F = 2203 N .......Ans
Perpendicular to the line:
Fnet = F - m g cos35 = 0
F = 7x 9.8 x cos35 = 56.2 N .....Ans
Angle relative to line of motion,
theta = tan^-1(56.2 / 2203) = 1.5 deg ..Ans
angle relative to horizontal = 35+1.5 = 36.5 deg .....Ans
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