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That's Mr. Idontknowhwhatimdoing to you sir!
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Velocity in x direction = v0cos(53) Velocity in y direction = v0sin(53)
Use d = d0 + v0t + (1/2)at^2 In the x direction: x = x0 + v0t + (1/2)at^2 solve for t Plug t into: y = y0 + v0t + (1/2)at^2 y - 3.05 = The distance by which the ball will clear, or fall short of, the bar.
Ok, starting from 0.8=25t-(1/2)(25/t)t^2 0.8 = 25t - (25t^2/2t) 0.8 = 25t - 12.5t 0.8 = 12.5t t = 0.8/12.5 Plug that back into our first equation: velocity = acceleration * time a = v / t a = 25m/s / (0.8/12.5) a / 9.8 = g force Plug in values and and
Careful, this is a torque problem. There are three sources of torque we need to solve for. The torque the person applies on the ladder. The torque the ladder applies on the wall, and the torque the wall exerts on the ladder. The torque exerted by the
Since we have tangential and radial acceleration our total acceleration is going to be: a = sqrt(tangential^2 + radial^2) Since the car doesn't slip or skid the total force on the car is zero: fmg = ma
a) Truck --> 40 = x0 + v0t + (1/2)at^2 40 = (1/2)at^2 t = sqrt(80/2.1) b) Car x = x0 + v0t + (1/2)at^2 x0 = (1/2)(3.4)t^2 - 40 c) Car v = v0 + at --> v = 3.4t Truck v = v0 + at --> v = 2.1t d) Plug in values and prosper.
1 yard = 3 feet 10.1666667yards * 3 = feet
We know that at the top of the motion the velocity of the ball is 0. Therefore 0 = v + at 0 = 19.6 + (-9.8)t t = 19.6/9.8
a = 9.8 m/s^2 x = x0 + v0*t + (1/2)*a*t^2 x = 0 + 0 + (1/2)*(9.8m/s^2)*(3^2)
4) x=(1/2)a*t^2 t=sqrt(2x/a) plug t into --> v=a*t 6) 90km/h = 25m/s v=at a=v/t --> a=25/t plug in a and solve --> x=x+vt+(1/2)at^2 0.8 = 25t - (1/2)(25/t)t^2 That gives you t which you can then plug back into a=v/t to get acceleration. You divide that by
