The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer:
False.
Explanation:
Decibels (dB) measure sound levels
As you coast down a long hill on your bicycle, potential energy from your height is converted to kinetic energy as you and your bike are pulled downward by gravity along the slope of the hill. While there is air resistance and friction slowing you down by a little bit, your speed increases gradually until you apply the brakes, causing enough friction to slow yourself and the bike to a stop at the bottom.
A roller coaster will have higher kinetic energy at the lower hill because it will have already been moving as opposed to the initial hill. But I'm not one hundred percent certain. You can always google this stuff, but I do know for sure that at the first hill, the roller coaster will have higher potential energy.
Hope this helps!
