Physics, 26.09.2019 20:00 jaliyahrobinson1
An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminus0.9tsquaredplus36t meters in t seconds. a. determine the velocity v of the object after t seconds. b. when does the object reach its highest point? c. what is the height of the object at the highest point? d. when does the object strike the ground? e. with what velocity does the object strike the ground? f. on what intervals is the speed increasing?
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What is the electric force acting between two charges of -0.0045 c and -0.0025 c that are 0.0060 m apart? use fe=kq1q2/r^2 and k = 9.00 x 10^9 n*m^2/c^2 a. 1.7 x 10^7 n b. -1.7 x 10^7 n c. -2.8 x 10^9 n d. 2.8 x 10^9 n
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Aparticle's position is given by x = 3.00 - 9.00t + 3t2, in which x is in meters and t is in seconds. (a) what is its velocity at t = 1 s? (b) is it moving in the positive or negative direction of x just then? (c) what is its speed just then? (d) is the speed increasing or decreasing just then? (try answering the next two questions without further calculation.) (e) is there ever an instant when the velocity is zero? if so, give the time t; if not, answer "0". (f) is there a time after t = 3 s when the particle is moving in the negative direction of x? if so, give the time t; if not, answer "0".
Answers: 3
An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reac...
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