The position of the particle is given by:
x(t) = t³ - 12t² + 21t - 9
Differentiate x(t) with respect to t to find the velocity x'(t):
x'(t) = 3t² - 24t + 21
Differentiate x'(t) with respect to t to find the acceleration x''(t):
x''(t) = 6t - 24
Hello
The word for Tepertaure is Celsius in the Metric System
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An object undergoing <span>uniform circular motion </span>is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction. So I'm thinking velocity
Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>
Answer: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2
Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.
Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:
acceleration in a circular movement= ω^2*r
so
a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2
a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2
