10. A tennis ball rolled off the edge of a table that has a height of 1,00 m.
1 answer:
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Answer:
A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water
B )angular frequency ( w ) =
c ) h = 1.34 m ( 4 / 3g )
Explanation:
A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density
B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables
volume of cone = hA /3.
h = height of cone, A = Base Area.
therefore the total volume of the Buoy above water level ( at rest )
= ( ) * 2 =
The part of the buoy immersed in water = x
then the net upward force will be
fb ( force of Buoy ) = fg ( force of gravity )
note force = Mass * Acceleration
force of buoy = * a = ( force of gravity )
therefore a =
angular frequency ( w ) =
C ) height of the each cone
therefore w = 2
back to the angular frequency :
therefore h = 1.34 m ( 4 / 3g )
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s