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3 years ago

Determine the amount of work you would need to do in order to stop a 1100kg car with 1400J of kinetic energy.

Physics

1 answer:

LiRa [457]3 years ago

6 0

Answer:

W = -1400 J

Explanation:

Let's use the relationship between work and kinetic energy

W = ΔK

kinetic energy is

K = ½ m v²

therefore in this case

K₀ = 1400 J

Thus

W = 0 -1400

W = -1400 J

The negative sign indicates that the work is done against the energy, that is, in the opposite direction to the movement

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A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

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Answer:

Energy stored in the capacitor is $U=2.7\times 10^{-5}\ J$

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At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

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Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

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Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

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