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3 years ago

One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of

Physics

1 answer:

Neko [114]3 years ago

5 0

Complete question is;

One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of 4.0 x 10-3m2 is kept in steam at 100 ˚C while the other end is in melting ice in a well-lagged container. The ice melts at a steady rate of 5.5 x 10- 4 kgs-1 and the thermal conductivity of the material of the bar is 401 Wm-1K-1. Calculate the specific latent heat of fusion of ice

Answer:

Specific Latent heat of fusion;

L_f = 13.6 × 10^(5) J/kg

Explanation:

We are given;

Length of bar; L = 0.8 m

Area;A = 4 × 10^(-3) m²

Temperature;ΔT= 100°C = 100 + 273 = 373 K

Rate of melting;m/t = 5.5 × 10^(-4) kg/s

Thermal conductivity;k = 401 W/m·K

Latent heat of fusion has a formula;

ΔQ/Δt = (m/t)•L_f

So, L_f = (ΔQ/Δt)/(m/t) - - - (1)

We also know that ;

ΔQ/Δt = (ΔT × k × A)/L

Plugging in the relevant values, we have;

ΔQ/Δt = (373 × 401 × 4 × 10^(-3))/0.8

ΔQ/Δt = 747.865 J/S

Plugging this value for ΔQ/Δt in equation 1 gives;

L_f = 747.865/(5.5 × 10^(-4))

L_f = 13.6 × 10^(5) J/kg

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3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

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Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :