Three identical 6.67-kg masses hang at rest from three identical springs, numbered as in the figure. Each spring has an unstretc
hed length of 0.23 m and a spring constant of 8.13 kN/m. How long, L , is spring 2?
[Hint: Draw some FBDs and apply Newton's second law. Remember to account for both the spring's stretch and its unstretched length.]
[Hint: Draw some FBDs and apply Newton's second law. Remember to account for both the spring's stretch and its unstretched length.]
1 answer:
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Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
For equipotential surface, dV = 0 so
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
Answer:
Therefore % increase in velocity is 18.23 %
Explanation:
we use the equality of mass flow rate and the areas
The percentage increase in velocity is
Δ v% = 100%
= .100%
= . 100%
= Therefore % increase in velocity is 18.23 %