Hooke's Law
F = k. Δx
Δx = 30 cm = 0.3 m
200 = k . 0.3
the spring stretch for 100 N:
The weight of a box is found to be 30 N. What is the approximate mass of the box?
2 answers:
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Since static friction is the minimum force required to just start the motion of a stationary object.
Here if we need to start an object from rest then we required F = 700 N
So for the first part of the above problem Force will be F = 700 N
Now if the box is already moving then we will have to use kinetic friction force between box and floor
now we can write the equation of net force as
here
now we will have
Answer:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by
where
E is the electric field
A is the cross-sectional area
is the angle between the direction of the electric field and the normal to A
The flux is maximum when , so we are in this situation and therefore
, so we can write
Here we have:
is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is
And so, we can find the magnitude of the electric field: