Force F= (2.84 i - 1.39 k)N acts on a pebble with position
vector r= (6.54...
Force F= (2.84 i - 1.39 k)N acts on a pebble with position
vector r= (6.54 j - 1.60 k)m , relative to the origin. What is the
resulting torque acting on the pebble about (a)
the origin and (b) a point with coordinates (4.96
m, 0, -1.43 m)?
An object moving in the xy-plane is subjected to the force F =
2xy i +...
An object moving in the xy-plane is subjected to the force F =
2xy i + 3y j N, where x and y are in m.
a. The particle moves from the origin to the point with
coordinates (a, b) by moving first along the x-axis to (a, 0), then
parallel to the y-axis. How much work does the force do?
b. The particle moves from the origin to the point with
coordinates (a, b) by moving first along the...
A) A man holds a 185-N ball in his hand, with the forearm
horizontal (see the...
A) A man holds a 185-N ball in his hand, with the forearm
horizontal (see the figure). He can support the ball in this
position because of the flexor muscle force , which is applied
perpendicular to the forearm. The forearm weighs 18.2 N and has a
center of gravity as indicated. Find (a) the magnitude of and the
(b) magnitude and (c) direction (as a positive angle
counterclockwise from horizontal) of the force applied by the upper
arm bone...
1)A 1.0 kg mass is located at (2.3, 0, -7.9) m and moving at (0,
5.7,...
1)A 1.0 kg mass is located at (2.3, 0, -7.9) m and moving at (0,
5.7, 3.9) m/s. What is the angular momentum of this mass about the
origin?
A.
(-3.0, 0.0, -76)
B.
(13, -9.0, 45)
C.
(45, -9.0, 44)
D.
(45, -9.0, 13)
E.
(-76, 0.0, -3.0)
F.
(0.0, 0.0, 0.0)
G.
(-13, 9.0, -45)
H.
(-45, 9.0, -13)
I.
(76, 0.0, 3.0)
J.
(-45, 9.0, -44)
K.
(-44, 9.0, -45)
L.
(44, -9.0, 45)
M.
(3.0, 0.0,...
1. Review your notes from 02-28 and your calculation of
the moment of inertia of the...
1. Review your notes from 02-28 and your calculation of
the moment of inertia of the "rigid rotator,"
two masses, 7 kg each
attached to a light titanium bar (neglect this weight)
separation 4.20 m
rotation axis halfway between the two masses, perpendicular to
the bar.
Our calculation of moment of inertia Iwas
I = ∑ left, right ( m r 2 ) = ( 7 k g ) ( 2.10 m ) 2 + ( 7 k g )...