A certain machine has a efficiency of
48% the velocity reatio of the machine
is 200 . Find the effort required to lift
load of 2 KN using this machine.
Answers
Given : A certain machine has a efficiency of 48%
the velocity ratio of the machine is 200 .
To Find : the effort required to lift load of 2 KN using this machine.
Solution:
Let say effort required to lift load of 2 kN using this machine. = x N
V.R.= 200. ( velocity ratio)
Hence Equivalent effort = 200x N
efficiency η = 48 %
Then (48/100)200x = 96x N
2 kN = 2000N
=> 96x = 2000
=> x = 2000/96
=> x = 125/6
=> x = 20.833 N
effort required to lift load of 2 KN using this machine. 20.833 N or 125/6 N
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Answer:
Solution
Step-by-step explanation:
Given : efficiency = 48%, V.R. = 200N, W= 2kN, P=?
efficiency = M.A / V.R × 100
48%= M.A / 200× 100
48%=M.A / 2
M.A= 48/ 2
[M.A = 24]
M.A = w/p
24 = 2 / p
p = 24 / 2
[ p = 12]
Effort required 12kN.