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2 years ago

The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?

Physics

1 answer:

RoseWind [281]2 years ago

5 0

Answer:

$E = 2.9775\times10^5 J$

Explanation:

Given: The latent heat of fusion for Aluminum is $L = 3.97\times10^5 J/Kg$

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

$E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J$

Therefore, energy required to melt 0.75 Kg aluminum

$E = 2.9775\times10^5 J$

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An incident ray of light strikes water at an angle of 30°. The index of refraction of air is 1.0003, and the index of refraction

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A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket

frutty [35]

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car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

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3 years ago