Acar of mass 800 kg is travelling at 30 m/s. the car then brakes suddenly and stops with a braking distance of 75 m find the work done on the car by the brakes

-360 kJ

Explanation:

Given:

m = 800 kg

v₀ = 30 m/s

v = 0 m/s

Δx = 75 m

Find: W

We can solve this using either forces or energy.

To use forces, first find the acceleration.

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2a (75 m)

a = -6 m/s²

Then apply Newton's second law:

∑F = ma

F = (800 kg) (-6 m/s²)

F = -4800 N

Work is force times distance:

W = FΔx

W = (-4800 N) (75 m)

W = -360,000 J

W = -360 kJ

If you want to use energy instead:

work = change in energy

W = ΔKE

W = ½mv² − ½mv₀²

W = ½ (800 kg) (0 m/s)² − ½ (800 kg) (30 m/s)²

W = -360,000 J

W = -360 kJ

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part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by