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Answer:
1.022 x 103 N.m
Explanation:
Solution
Given:
The weight of the block of mass m₂ is :
w₂ = m₂*g
Where
w₂ = 39 x 9.8 = 382.2 N
Then,
The weight of the block of mass m₁
w₁= m₁*g;
so,
w₁ = 12 x 9.8 = 117.5 N
Thus,
The tension wrapped in cord on drum (80 cm) T₁ = F - w₁
Now,
T₁ = 1200 - 117.5
T₁ = 1082.5 N
The tension wrapped in the cord on drum (41 cm) T₂ = w₂;
T₂ = 382.2 N
Hence,
We calculate net torque on the center of the drum:
The net torque = T₁ x 0.8 + T₂ x 0.41;
= 1082.5 x 0.8 + 382.2 x 0.41;
= 1.022 x 103 N.m
Therefore, the resulting torque applied to the system is 1.022 x 103 N.m
It is very large in size, I don’t have an exact number but i believe it is over the 1,000s.
A single pulley changes the direction of the effort, but it has
no mechanical advantage. The output force is the same as
the input force, so we'd say that the mechanical advantage is 1.
If there are two pulleys, with the rope going up and down and
around between them several times on its way from the effort
to the load, an arrangement that I think is called 'block and tackle',
then the mechanical advantage turns out to be the number of
strands of rope that are supporting the load.