If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m
Answer:
32000joule.
Explanation:
given
mass. (m)=160kg
speed (v)=20m/s
now
kinetic energy =1/2 (mv²)=1/2 ×{160×20²}=32000joule.
A) 
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:
Using (2) we can rewrite this as
And using (1), we find
Substituting 
into the last equation, we find the value of x:
B) 
In this case, the kinetic energy is 1/10 of the total energy:
Since we have
we can write
And so we find:
Answer:
A. kinetic energy
B. angular velocity
E. angular position
Explanation:
The quantities that cannot be constant if a constant net torque is exerted on an objecta are:
A. Kinetic energy. If a torque is applied, the linear or angular speed will be changing at a rate proportional to the torque, so the kinetic energy will change too.
B. Angular velocity. It will change at a rate equal to the torque.
C. Angular position. If the angular velocity changes, the angular position will change.
A) use v=u+at for both
First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.
b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.
First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
