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tamaranim1 [39]
3 years ago
10

a 2000 kg car moving down the road runs into a 5000 kg stationary suv. The car applies a force of 1400 n on the suv what is the

magnitude of force applied by the suv on the car
Physics
1 answer:
masha68 [24]3 years ago
3 0

Answer:

F_suv= 49050 N

Explanation:

We are told that a 2000 kg car moving down the road runs into a 5000 kg stationary suv. The car applies a force of 1400 N on the suv.

Now, according Newton's first law of motion, an object will continue in it's present state of rest except it is acted upon by an external body.

This means the force acting on the stationary Suv is force of gravity.

Thus; F_suv = 5000 × 9.81

F_suv= 49050 N

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Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

          ΔH is enthalpy change

          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

           Entropy change,ΔS = 600 J/K

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  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

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                                     = 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

                                                   = 800-491.891

                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

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