Answer
given,
mass of the ball = 3 kg
swing in vertical circle with radius = 2 m
work done by the gravity = ?
work done by the tension = ?
Work done by the gravity = - m g Δh
Δ h = 2 + 2 = 4 m
Work done by the gravity =
= -117.6 J
work done by gravity is equal to -117.6 J
Work done by tension will be equal to zero.
Zero because tension is always perpendicular to velocity
work done by tension is equal to 0 J
A car A house A phone they all can be renewable
Answer:
a = the lowest critical speed of the shaft 882.81 rad/s
b = new diameter 0.05m or 50mm
c = critical speed 1765.62rad/s
Explanation:
see the attached file
Answer:
Scalar product is between ║A║ ║ B║ and -║A║ ║ B║
Explanation:
Dot product between vec A and vec B is
A.B = ║A║ ║ B║ cos θ
Here, both ║A║ and ║B║ are positive and value of cos θ depends upon θ and lies between 1 and -1
So, Scalar product is between ║A║ ║ B║ and -║A║ ║ B║
Answer:
a) W = 643.5 J, b) W = -427.4 J
Explanation:
a) Work is defined by
W = F. x = F x cos θ
in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero
W = F x
let's calculate
W = 165 3.9
W = 643.5 J
b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes
sin θ = Wₓ / W
cos θ = Wy / W
Wₓ = W sinθ = mg sin θ
Wy = W cos θ
the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero
W = Wₓ x cos 180
W = - mg sin 34 x
let's calculate
W = -20 9.8 sin 34 3.9
W = -427.4 J
The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.
