Answer:
α = 395 rad/s²
Explanation:
Main features of uniformly accelerated circular motion
A body performs a uniformly accelerated circular motion when its trajectory is a circle and its angular acceleration is constant (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.
There is tangential acceleration (at) and is constant.
at = α*R Formula (1)
where
α is the angular acceleration
R is the radius of the circular path
There is normal or centripetal acceleration that determines the change in direction of the velocity vector.
Data
R = 0.0600 m :blade radius
at = 23.7 m/s² : tangential acceleration of the blades
Angular acceleration of the blades (α)
We replace data in the formula (1)
at = α*R
23.7 = α*(0.06)
α = (23.7) / (0.06)
α = 395 rad/s²
Answer:
false
Explanation:
I think I am right with this
Answer:
a' =4.15 m/s²
Explanation:
Given that
m= 3.2 kg
F₁ = 1.9 i −1.9 j N
F₂=3.8 i −10.1 j N
From second law of Newton's
F(net) = m a
F₁ + F₂ = m x a
1.9 i −1.9 j + 3.8 i −10.1 j = 3.2 a
a = 1.78 i - 3.75 j m/s²
The resultant acceleration a'

a' =4.15 m/s²
First put the speed in m/s. 120km/h = 33.33m/s. Now the position function is the integral of velocity, and velocity is in turn the integral of acceleration. The velocity is:

Now we integrate this expression to get the position. The constant of integration will be the distance the truck travels.

Here we set the distance, 35m as negative because I assumed the stopping point of the truck is the origin. Putting t=0 shows it starts at -35m.
Now solve the following equation for the time, t using the quadratic equation:

and choose the value t=1.149s