An iceberg at its melting point

Physics

1 answer:

iren2701 [21]2 years ago

5 0

The mass of the iceberg is 71.9 kg

Explanation:

The amount of thermal energy needed to completely melt a substance at its melting point is given by

$Q=\lambda m$

where

$\lambda$ is the latent heat of fusion

m is the mass of the substance

In this problem, we have a block of ice at its melting point (zero degrees). The amount of heat given to the block is

$Q=2.40\cdot 10^7 J$

And the latent heat of fusion of ice is

$\lambda = 334 J/g$

So, we can re-arrange the equation to find m, the amount of ice that will melt:

$m=\frac{Q}{\lambda}=\frac{2.40\cdot 10^7}{334}=71856 g = 71.9 kg$

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

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A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

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Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$