55. Calculate the density of a cubical ice block of side 50
cm if a force 1125 N applied to it produces an accel-
eration of 10 m s-2 in it. Neglect the force of friction.
Assume the ice block remains in solid state without
melting
Answers
Answered by
9
Answer:
given : edge of cube a = 50 cm
f = 1125 N, a = 10m/sec²
f= ma
1125 = m*10
therefore m = 1125/10
=> 112.5 kg
density = m/unit volume
volume of the block = a³
=> 50³ = 125000 cu cm
therefore density of ice cube = 112.5/125000
=> 0.0009kg/cc
=> 0.0009*1000gm/cc
=> 0.9 gm/cc
Answered by
0
Answer:
0.9 g cm^3 will be the density
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