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Nina [5.8K]
3 years ago
11

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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Given that,

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We know that,

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1 h = 3600 s

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v=891.66\ km/h =\dfrac{891.66\times 1000\ m}{3600\ s}\\\\v=247.68\ m/s

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3 years ago
Objects in space are moving at a constant velocity in a straight line.
tigry1 [53]

Answer:

C

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An object in motion will stay in motion unless acted on by a net positive or negative force.

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pavement is defined as the surface of Road or sidewalk.

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brainly.com/question/28456065

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2 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

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  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

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b)

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       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
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       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

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