The question seems to be incomplete. It is asking about a spring with a spring constant of 100 N/m, but it is not clear what the question is asking for. It could be asking about the displacement of the spring, the force applied to the spring, or the period of oscillation of the spring. Without more context, it is difficult to provide a specific answer. **

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When a spring has a spring constant of 15 N/cm, it means that it requires a force of 15 Newtons to stretch or compress the spring by 1 centimeter. The spring constant is a measure of the stiffness of the spring, and a higher spring constant indicates a stiffer spring that requires more force to deform. This value is important in understanding the behavior of the spring in various mechanical systems and can be used to calculate the amount of force needed to stretch or compress the spring by a certain distance. **

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A T-slot nut with a threaded screw and spring ball typically holds better in an aluminum profile compared to a hammer head nut with a screw. The spring ball provides additional tension and grip within the T-slot, ensuring a more secure connection. The spring ball acts as a detent mechanism, preventing the nut from accidentally sliding out of the T-slot during assembly or disassembly, making it a preferred choice for applications where vibration or movement may be present. **

A T-slot nut with threaded insert and spring ball generally holds better in an aluminum profile than a hammer head nut with a screw. The spring ball provides a secure grip and prevents the nut from slipping or rotating within the T-slot. The function of the spring ball is to apply pressure against the T-slot, creating a more secure and stable connection between the nut and the aluminum profile. This makes it a more reliable and effective option for holding in an aluminum profile compared to a hammer head nut with a screw. **

To calculate the elongation of the spring, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the elongation of the spring. The formula is F = kx, where F is the force applied, k is the spring constant, and x is the elongation of the spring. Rearranging the formula to solve for x, we get x = F/k. Plugging in the values, x = 370 N / (70 N/cm) = 5.29 cm. Therefore, the spring elongates by 5.29 cm when pulled with a force of 370 N. **