Answer:
vx = 65 yd/3 sec = 21.7 yd/sec since horizontal speed is constant
vy = g t = (32 / 3) yd/sec^2 * 1.5 sec = 16 yd/sec where 32/3 is the acceleration due to gravity in yds / sec^2 and 1.5 is the time to travel each way in the vertical direction
V = (vx^2 + vy^2)^1/2 = (21.7^2 + 16^2)^1/2 = 27 yd/sec
tan theta = vy/vx = 16 / 21.7 = .737 theta = 36.4 deg
You can check using the range formula:
R = v^2 sin (2 theta) / g = 27^2 * .955 / (32 / 3) = 65.3 yds
The difference from 65 yds may be rounding error.
The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry
The formula used for finding the tangential speed (speed of something that is moving in a circular path) of an orbiting object is:
V₍t₎ = ωr
V₍t₎ = tangential speed or velocity
ω = angular velocity
r = radius of the circular path
if time taken t is only given then use this formula to calculate the tangential speed:
V₍t₎ = 2πr/t, t is time taken
I dont know what you are trying to say but okay
Answer:An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.People sometimes think that objects must stick together in an inelastic collision. However, objects only stick together during a perfectly inelastic collision. Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved.
hope this helps have a nice day❤️
Explanation:
