Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right
Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s
Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right
To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:



The angular displacement is given as the form:
In the equlibrium we have to
and in the given position we have to

Derived the expression we will have the equivalent to angular velocity

Replacing,

Finally

Therefore the maximum angular displacement is 9.848°
Answer:
he factor for the temporal part 1.296 107 s² = h²
m / s² = 12960 km / h²
Explanation:
This is a unit conversion exercise.
In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.
In this case, it is requested to convert a measure km/h²
Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is m / s² , which is a measure of acceleration. Let's cut this down
the factor for the distance is 1000 m = 1 km
the factor for time is 3600 s = 1 h
let's make the conversion
m / s² (1km / 1000 m) (3600 s / 1h)²
note that as time is squared the conversion factor is also squared
m / s² = 12960 km / h²
the factor for the temporal part 1.29 107 s² = h²
Answer:
The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.
Explanation:
This is a horizontal bar.
For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.
Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).
Answer:
The velocity of the collar will be 3.076 ft/s
Explanation:
Given data
weight of the disk, Wa = 20lb
weight of rod BC, Wbc = 4lb
weight of collar, Wc = 1lb
Considering the equation of equilibrium
Vb = 1.5Wbc
Wa = 1.875 Wbc
to calculate the velocity of the collar using energy conservation equation
T1 + V1 = T2 + V2
=>
=>
=>
Wbc = 1.18rad/sec
i.e.

= 3.076 ft/ s