The lightest duty and most widely used non flexible metal conduit is
1 answer:
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Answer:
a. 30 N / m
b. 9.0 N
Explanation:
Given that
Unstretched length of the spring, = 20.0cm = 0.2m
a) When the mass of 4.5N is hanging from the second spring, then extended length Is
= 35.0cm = 0.35m
So, the change in spring length when mass hangs is
= (0.35 - 0.20) m
= 0.15m
As spring are identical
Let us assume that the spring constant be "k", so at equilibrium
Restoring Force on spring = Block weightage
kx = W = 4.50
= 30 N / m
b) Now for the third spring, stretched the length of spring is
= 50cm = 0.5m
So, the change in spring length is
= (0.5-0.20)m
= 0.30m
At equilibrium,
Restoring Force on spring = Block weightage
Now using all mentioned and computed values in above,
= 30(0.3)
= 9.0 N
Answer:
to the right.
to in the upwards direction.
Explanation:
In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).
So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:
F=ma
so in order to find the components of the force, we need to start by finding its acceleration.
Acceleration is found by using the following formula:
so we can subtract the two vectors, like this:
which yields:
or:
so now I can find the components of the force:
which yields:
F=(2.31i+2.1j)N
so the components of the force are:
to the right.
to in the upwards direction.
