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svp [43]
3 years ago
11

A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed

in the water. What is the final water temperature? State the assumptions you need to make in your calculations.
[Given the value of specific heat capacity of water is 4200 J kg^-1 °C^-1 and that of iron is
450 J kg^-1 °C^-1]​
Physics
2 answers:
d1i1m1o1n [39]3 years ago
4 0

Answer:

Assuming there is no heat loss to the surrounding.

Heat lost by iron equals heat gained by water.

0.2(450)(50-x)=0.2(4200)(x-30)

x=31.94 °C

Explanation:

jeka943 years ago
3 0
Your answer would be 39.94 c
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you are hiking along a river and see a tall tree on thhe opposite bank. You measure the angle of elevation of the top of the tre
Sidana [21]

Answer:

Explanation:

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A constant tone is being applied to a speaker. The voltage across the speaker is 5 volts. The voltage across the speaker is incr
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5 0
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Four velcro-lined air-hockey disks collide with each other in a perfectly
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Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

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The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, \Sigma P_{initial} is given as follows;

\Sigma P_{initial} = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ \Sigma P_{initial} = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ \Sigma P_{initial} = -20·i + 130·j

By the law of conservation of linear momentum, we have;

\Sigma P_{initial} = \Sigma P _{final} = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

\Sigma P _{final} = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

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\left | v \right | ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0
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