A leaf spring in an off-road truck with a spring constant k of 87.6 kN/m (87,600 N/m) is compressed a distance of 6.2 cm (0.062 m) from its original unstretched position. What is the increase in potential energy of the spring (in kJ)?

0.00367 kJ

Explanation:

Using,

ΔEp = 1/2ke² Equation 1

Where ΔEp = increase in potential energy, k = spring constant of spring, e = extension/compression

Given: k = 87600 N/m, e = 0.0062 m

Substitute into equation 1

ΔEP = 1/2(87600)(0.0062²)

ΔEP = 3.367 J.

ΔEp = 0.00367 kJ

Hence the change in potential energy of the spring =  0.00367 kJ

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c.arachnids

  scorpions are animals in the order scorpiones, under the class arachnida, which makes them a distant cousin of spiders

happy to

pls mark as brainliest.

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .