Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.47 kg mass to the spring's other end. if the spring stretches 3.1 cm from its equilibrium position, what is the spring constant? 133 incorrect: your answer is incorrect. n/m

The force needed to extend or compress a spring is directly proportion to the applied stress in spring. The spring constant of a given spring is 148.71 N/m.

The Hook's law:

Within the elastic limit the force needed to extend or compress a spring is directly proportion to the applied stress in spring.

The force applied by the wight

Where,

m = mass = 0.47 kg

g - gravitational acceleration =

Put the values in the equation,

Spring constant can be calculated using the Hook's law,

Therefore, the spring constant of a given spring is 148.71 N/m.

To know more about Hook's law,

link

  148.6 N/m

Explanation:

The 0.47 kg mass applies a force of ...

   F = ma = (0.47 kg)(9.8 m/s^2) = 4.606 N

which stretches the spring by 3.1 cm, or 0.031 m. That means the spring constant is ...

  (4.606 N)/(0.031 m) ≈ 148.6 N/m

Additional comment

The supplied numbers have 2 significant figures. If the spring constant is rounded to 2 significant figures, it would be 150 N/m.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.