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3 years ago

7

It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

ic energy is converted into heat via friction. Find the minimum speed of a lead bullet (initial temperature = 43.0° C) for such an event to happen? (Use Lf = 2.32 104 J/kg and melting point = 327.3° C.)

1 answer:

TEA [102]3 years ago

3 0

Answer:

$v=346.05\ m.s^{-1}$

Explanation:

Given:

initial temperature of the lead bullet, $T_i=43^{\circ}C$

latent heat of fusion of lead, $L_f=2.32\times 10^4\ J.kg^{-1}$

melting point of lead, $T_m=327.3^{\circ}C$

We have:

specific heat capacity of lead, $c=129\ J.kg^{-1}.K^{-1}$

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

$\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)$

$\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f$

$\frac{1}{2} m.v^2=m(c.\Delta T+L_f)$

$\frac{v^2}{2} =129\times(327.3-43)+23200$

$v=346.05\ m.s^{-1}$

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3 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

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Part b)

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Explanation:

Part A)

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here we have

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so power required to melt is given by

$P = \frac{Q}{t}$

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Part B)

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3 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

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As per the question:

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Length of the antenna, l = 1.76 m

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Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

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3 years ago