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2 years ago

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A blacksmith heats a 1.5 kg iron horseshoe to 500 C, containing 20 kg of water at 18 C then plunges it into a bucket What is the

equilbrlum temperature? Express your answer using two significant figures.

1 answer:

Tema [17]2 years ago

8 0

Answer:

21.85 C

Explanation:

mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C

mass of water = 20 kg, initial temperature of water, T2 = 18 C

let T be the equilibrium temperature.

Specific heat of iron = 449 J/kg C

specific heat of water = 4186 J/kg C

Use the principle of caloriemetry

heat lost by the hot body = heat gained by the cold body

mass of iron x specific heat of iron x decrease in temperature = mass of water x specific heat of water x increase in temperature

1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)

336750 - 673.5 T = 83720 T - 1506960

1843710 = 84393.5 T

T = 21.85 C

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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

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2 years ago

Is distance traveled by a moving train qualitative discrete continuous or na?

butalik [34]

Continuous. Discrete values are values like 1, 2, 3, 4, etc. - they're values that are <em>distinct</em>, and typically there's some idea of a <em>next </em>and a <em>previous </em>value. When we're counting whole numbers, there's a definitive answer to which number comes after, and which number comes before. With continuous values, there's no real "next" or "last" value.

Motion is measured with <em>continuous </em>values; a train might move 300 yards in 1 minute, but we can look at smaller and smaller chunks of time to keep getting shorter and shorter distances. There is no <em />"next" distance the train moves after those 300 yards - it just doesn't make sense for there to be.

It's also measured <em>quantitatively</em>, not <em>qualitatively</em>. This just means that we can use numerical values to measure it, rather than other descriptors like color, smell, or taste.

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2 years ago

If m = 2,000, p = 2. 25, and y= 6,000, what is velocity?

Karo-lina-s [1.5K]

The velocity is 6.75

The velocity in the equation stated above can be calculated as follows

m= 2,000
p= 2.25

y= 6000

velocity= 2.25 × 6000/ 2000
= 13500/2000
= 6.75

Hence the velocity is 6.75

Please see the link below for more information.
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1 year ago

En una fundición hay un horno eléctrico con capacidad para fundir totalmente 540 kg de cobre. Si la temperatura inicial del cobr

11Alexandr11 [23.1K]

Answer:

Q = 2.95*10^5 kJ

Explanation:

In order to calculate the energy required to melt the cooper, you first calculate the energy required to reach the boiling temperature. You use the following formula:

$Q_1=mc(T_b-T_1)$ (1)

m: mass of cooper = 540 kg

c: specific heat of cooper = 390 J/kg°C

Tb: boiling temperature of cooper = 1080°C

T1: initial temperature of cooper = 20°C

You replace the values of the parameters in the equation (1):

$Q_1=(540kg)(390\frac{J}{kg.\°C})(1080\°C-20\°C)=2.23*10^8J$

Next, you calculate the energy required to melt the cooper by using the following formula:

$Q_2=mL_f$ (2)

Lf: melting constant of cooper = 134000J/kg

$Q_2=(540kg)(134000\frac{J}{kg})=7.24*10^7J$

Finally, the total amount of energy required to melt the cooper from a temperature of 20°C is the sum of Q1 and Q2:

$Q=Q_1+Q_2=2.23*10^8J+7.24*10^7J=2.95*10^8J=2.95*10^5kJ$

The total energy required is 2.95*10^5 kJ

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2 years ago

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

yawa3891 [41]

Answer:

Explanation:

Time taken to complete one revolution is called time period.

So, Time period, T = 1 s

Diameter = 1.6 mm

radius, r = 0.8 mm

Let the angular speed is ω.

The relation between angular velocity and the time period is

$\omega =\frac{2\pi}{T}$

ω = 2 x 3.14 = 6.28 rad/s

The relation between the linear velocity and the angular velocity is

v = r x ω

v = 0.8 x 10^-3 x 6.28

v = 0.005 m/s

6 0

2 years ago