Initial inertia = 3.6 kg . m^2
final inertia = 0.80 kg.m^2rotational velocity final = 4.1 rev/s
rotational velocity initial = 0.90 rev/s
lo= lf(final inertia) * Wf(rotational velocity final)/ Wo (rotational velocity initial)
lo = 0.8*4.1 / 0.90 = 3.6 kg m2
Answer:
Wl = 1740 N
Explanation:
maximum lift weight unaided = force exerted (F) = 650 N
length of the wheelbarrow (L) = 1.4 m
weight of the wheelbarrow (w) = 80 N
distance of center of gravity of the wheel barrow from the wheel = 0.5 m
distance of center of gravity of the load from the wheel = 0.5 m
find the weight of the load (Wl)
from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive
ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0
(F x 1.4) = ((Wl x 0.5) + (w x 0.5)
Wl =
Wl =
Wl = 1740 N
Answer:
-929.5Joules
Explanation:
To get the work done by sam, we will calculate the kinetic energy of sam expressed as;
KE = 1/2mv²
m is the mass = 1100kg
v is the velocity = 1.3m/s
KE = 1/2(1100)(1.3)²
KE = 550(1.69)
KE = 929.5Joules
Since Sam is opposing the direction of movement, work done by him will be a negative work i.e -929.5Joules
Here a cat is running at constant speed which is given as 10 km/h for 5s
So here the average speed is defined as total distance moved in total time interval
so here it is given by
since
here speed of cat is constant so it will remain the same
And hence the average speed and instantaneous speed at any instant for this duration will remain the same
so here answer would be
<em>average speed = 10 km/h</em>
<em>instantaneous speed = 10 km/h</em>
