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Thepotemich [5.8K]
1 year ago
6

A blacksmith heats a 1,540 g iron horseshoe to a temperature of 1445°c before dropping it into 4,280 g of water at 23.1°c. if th

e specific heat of iron is 0.4494 j / g °c, and the water absorbs 947,000 j of energy from the horseshoe, what is the final temperature of the horseshoe-water system after mixing
Physics
1 answer:
Marta_Voda [28]1 year ago
3 0
Given:
m₁ = 1540 g, mass of iron horseshoe
T₁ = 1445 °C, initial temperature of horseshoe
c₁ = 0.4494 J/(g-°C), specific heat

m₂ = 4280 g, mass of water
T₂ = 23.1 C, initial temperature of water
c₂ = 4.18 J/(g-°C), specific heat of water
L = 947,000 J heat absorbed by the water.

Let the final temperature be T °C.
For energy balance,
m₁c₁(T₁ - T) = m₂c₂(T - T₂) + L
(1540 g)*(0.4494 J/(g-C))*(1445-T C) = (4280 g)*(4.18 J/(g-C))*(T-23.1 C) + 947000 J
692.076(1445 - T) = 17890(T - 23.1) + 947000
10⁶ - 692.076T = 17890T - 413259 + 947000
466259 = 18582.076T
T = 25.09 °C

Answer: 25.1 °C
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A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson
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option D

Explanation:

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so, here n = 3

now,

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f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

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An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?
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The answer is -15.625m/s².


Acceleration is the change in velocity over a period of time. It can be computed using the formula:


a = \dfrac{vf-vi}{t}

Where:

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Now let's see what was given in your problem:

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But look at the problem, it shows no time. We need to solve for time from the time it moved till it reached the red light 20 m away.


Time can be computed using the kinematics formula:

d = \dfrac{vi+vf}{2} *t


We just derive the formula from the equation by filling out what we know first. 

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a = -15.625m/s^{2}


Notice that the acceleration is negative. This means that the car decelerated or slowed down.

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