Answer:
5.35 rad/s
Explanation:
From the question, we are toldthat an Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick.
Solving this question, the potential energy of the particles must equal to the Kinectic energy i.e
P.E=K.E
Mgh= m½Iω²-------------eqn(*)
Where M= mass of the particles
g= acceleration due to gravity= 9.81m/s^2
ω= angular speed =?
h= height of the particles in the stick on the metre stick= ( 50cm + 80cm)= (0.5m + 0.8m)= 130cm=1.3m
If we substitute the values into eqn(*) we have
m×9.81× (1.3m)= 1/2× m×[ (0.5m)² + [(0.8m)²]× ω²
m(12.74m²/s²)= 1/2× m× (0.25+0.64)× ω
m(12.74m²/s²)= 1/2× m× 0.89× ω²
We can cancel out "m"
12.74= 1/2×0.89 × ω²
12.74×2= 0.89ω²
25.48= 0.89ω²
ω²= 28.629
ω= √28.629
ω=5.35 rad/s
Hence, the angular speed of the meter stick as it swings through its lowest position is 5.35 rad/s
Answer:
So -24.5 m merely means the depth of the base of the tower ie the ground from the top of the tower. So wrt the ground height of the tower = 24.5 m.
Explanation:
T² caries directly as R³ .
This is Kepler's 3rd law of planetary motion .
Answer:
re believed to influence the Earth's magnetic field. ... As heat is transferred outward toward the mantle, the net trend is for the inner boundary of the liquid region to freeze, causing the solid inner core to grow
Explanation:
