A common technique used to measure the force constant k of a spring is the following: Hang the spring vertically, then allow a mass m to stretch it a distance d from the equilibrium position under the action of the “load” m g.

Solution and Explanation:

here, as per the given data

the weight suspended from the spring , W is equals to 98 N

the compression of spring , d is equals to 51 m

now, let the spring constant be K

Now, equating the forces vertically

K multiply with d is equals to W

K multiply with 51 is equals to 98

Now, we have to solve for K

K = 1.92 N/m

Therefore, the spring constant is 1.92 N/m

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop