an object of mass 500 gram is dropped from a tower of height 5m calculate its Momentum when it touches the ground given the is equal to 10 M per second square
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v²=u²+2as
u=0
s=5m
a=10m/s²
v²= 0+2(10)(5)
v²= 100
v= 10m/s
momentum = mass × velocity
m=500g
m=0.5kg
p=0.5×10
p= 5kg m/s
u=0
s=5m
a=10m/s²
v²= 0+2(10)(5)
v²= 100
v= 10m/s
momentum = mass × velocity
m=500g
m=0.5kg
p=0.5×10
p= 5kg m/s
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Answered by
1
Explanation:
mass = 500 gram
mass in kg will be 0.5 kg
height = 5 m
acceleration due to gravity = 10 m/s²
u = 0m/s
By using suitable formula velocity will be
v² - u² = 2as
v² = 2×5×10 + 0×0
v² = 100
v = 10 m/s
Hence,velocity = 10 m/s
momentum = velocity × mass
momentum = 10 × 0.5
momentum = 5 kg m/s
Hence,momentum = 5 kg m/sec.
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