at 4s, the speed is 24 + 4g at 5s, the speed is 24 + 5g so the average speed over the interval (for one second) is 24 + 4.5g the distance is the speed times the time
h = .5 * g * t^2 34 = .5 * 9.8 * t^2
constant velocity means no acceleration, which means zero net force so the third force is equal in magnitude to the resultant of the given forces; but opposite in direction the magnitude of the resultant of the given forces is __ sqrt(61.9^2 + 52.0^2) the tangent of the angle ...
the work done on the car is reflected by the change in kinetic energy w = [.5 * 2000 * 20^2] - [.5 * 2000 * 10^2] the units are joules
the kinetic energy at the bottom is the kinetic energy at the top plus the gravitational potential across the circle .5 m (Vb)^2 = [.5 m (Vt)^2] + [m g h] dividing by .5 m (Vb)^2 = (Vt)^2 + 2 g h v^2 = 3^2 + [2 * 9.8 * (2 * .8)]
n = d - 4 [(d - 4) / d] + [d / (d - 4)] = 5/2 multiplying by [2 * d * (d - 4)] [2 (d - 4)^2] + (2 d^2) = 5 d (d - 4) 2d^2 - 16d + 32 + 2d^2 = 5d^2 - 20d 0 = d^2 - 4d - 32 factoring __ 0 = (d - 8)(d + 4) 0 = d - 8 __ 8 = d
this is like a projectile problem turned sideways 0 = -(.5 * 3.7 * t^2) + (30 * t) solve the quadratic for t (you want the non-zero solution)
the crate is sliding, so μk (the kinetic coefficient) is the relevant one the direction of the frictional force is OPPOSITE to the direction of motion the magnitude of the force is: the normal force of the crate on the ramp [m * g * cos(30º)]; multiplied by the coeff...
the force diagram reflects the physical setup the walker is 15.0 m from either end; and is displacing the wire 1.0 m this means that the vertical component of the tension is one-fifteenth the tension in the wire (on BOTH sides) so the force exerted by the walker is 3675/15 * 2...
because it starts from rest and accelerates uniformly, the average velocity is 1/2 the final velocity convert the km/h into m/s [multiply by 1000/3600] the distance is the average velocity, multiplied by the time (8.5 s)
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