The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:
where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:
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The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t =
, at = v, so
Vertical columns on the periodic table are called groups.
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Answer:
0.2631 N/C
Explanation:
Given that:
The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m
The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m
The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons
Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)
= 0.0006
The number of electron flow per second is calculated as follows:
The magnitude of the electric field is:
E = 
E = 
E = 
E = 0.2631 N/C
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