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Masteriza [31]
2 years ago
15

A bottle of water with mass 0.9 kg is left out in the sun, the radiation from the sun warms up the water bottle. If the water bo

ttle was initially 10°C and the sun provided 120,000 J of thermal energy, calculate the final temperature of the water. Use C = 4182 j/kgK for the specific heat of water.
Physics
1 answer:
natita [175]2 years ago
8 0

Answer:

Final temperature, T2 = 314.9 Kelvin

Explanation:

Given the following data:

Mass = 0.9kg

Initial temperature, T1 = 10°C to Kelvin = 10 + 273 = 283K

Quantity of heat = 120,000 J

Specific heat capacity = 4182 j/kgK

To find the final temperature;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making dt the subject of formula, we have;

dt = \frac {Q}{mc}

Substituting into the equation, we have;

dt = \frac {120000}{0.9*4182}

dt = \frac {120000}{3763.8}

dt = 31.9K

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 31.9 + 283

T2 = 314.9 Kelvin

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dangina [55]
When an object falls or is dropped from rest it's initial velocity is zero.
Using the equations for a motion in straight line. I can find the time it takes to reach 3.0 m down (half way).
x = vt - 4.9t²
-3 = 0 - 4.9t²
-3/-4.9 = t²
0.6122 = t²
0.7825 sec = t

v = v - gt
v = 0 - 9.8(0.7825)
v = -7.67 m/s
the negative denotes downward direction.

You  could also solve the problem using potential and kinetic energy.

Since it starts with maximum PE and gets converted to KE when it hits the ground. mgh = mv²/2
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7 0
3 years ago
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Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral
cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ      

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂)   (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

6 0
3 years ago
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8_murik_8 [283]

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Most of the time you are trying to use wire with the least resistance.

A meter of copper has a listed resistance of 0.024 ohms / meter. The wire is a 19 guage wire which makes it pretty thin.

===============

I'm not sure what you are asking. If want the resistance of something in terms of what would increase the resistance of the same material for both calculations then

Rule 1: It you decrease the wire diameter, you increase the resistance

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