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shepuryov [24]
2 years ago
7

Water of mass, m at 100 ℃ is added to 0.50 kg of water at 20 ℃ in a lagged calorimeter of thermal capacity 105 JK −1 . If the sp

ecific heat capacity of water is 4200 kg −1K −1 and the final temperature of the mixture is 70℃, determine the value of m.
Physics
1 answer:
lorasvet [3.4K]2 years ago
8 0

The value of m given that the temperature of the mixture is 70°C is : 0.875 Kg

<h3>Determine the value of m </h3>

Applying the principle of energy conservation

Heat lost by the hot body = heat absorbed by the cold water + calorimeter

= m * 4200 * ( 100 - 70  ) = 0.50 * 4200 * ( 70 -20 ) + 105 * ( 70 -20 )

= m * 126000 = 110250

therefore ;

m = 110250 / 126000

  = 0.875 kg

Hence we can conclude that The value of m given that the temperature of the mixture is 70°C is : 0.875 Kg

Learn more about specific heat capacity : brainly.com/question/10541586

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6.1328 kg

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The weight of the bowling ball will balance the buouyant force

W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg

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5 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h
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Answer

given,

time  = 10 s

ship's speed = 5 Km/h

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a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

a_1=\dfrac{1.39}{10}

a₁ = 0.139 m/s²

F₂ =m x a₂

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Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

a_3=\dfrac{5.282}{10}

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

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F₂/F₁

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4 years ago
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Answer:

1.01 × 10⁵ Pa  

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Hence, the total pressure is 1.01 × 10⁵ Pa.

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