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sveta [45]
4 years ago
9

How much energy is needed to melt 0.225 kg of lead note that the initial tempreture of the lead immediately after it has melted

is 27.3 degree celicius
Physics
1 answer:
iren [92.7K]4 years ago
3 0

Answer:

The energy needed to melt 0.225 kg of lead with an initial temperature of 27.3 °C is  15.1563 kJ

Explanation:

To solve the question, we are required to use the properties of the lead, and the in this case the necessary properties are the heat capacity to rise the temperature of the lead to the melting point temperature, then at the melting point temperature, the latent heat required to melt the lead

The melting point of lead T₂ is 327.5 °C

The latent heat of fusion of lead, L = 4.799 kJ/mol

The specific heat capacity of lead, c = 0.13J/g K

The molar mass of lead, M = 207.2 g

Initial temperature T₁ = 27.3 °C

Mass of lead = 0.255 kg

Change in temperature from initial temperature, to melting point temperature = (T₂ - T₁) = ΔT

Heat required to melt the lead = H

Note that the latent heat is given in kJ/mol therefore in the following equation in the calculation for latent heat component we have m/M

Therefore to melt the lead we have H = Sensible heat component (m×c×ΔT) + latent heat component (m/M×L)

Therefore to melt the lead we have H = m×c×ΔT + m/M×L

= 0.255 kg × (1000 g/kg) × (327.5 - 27.5)×0.13J/g K + ((0.225 kg × 1000 g/kg)/(207.2 g))× 4.799 kJ/mol × 1000J/ kJ

9945.0 J +5211.3 J = 15156.3 J

The heat energy required to melt 0.225 kg of lead that was initially at 27.3 degrees Celsius is  15156.3 J or 15.1563 kJ

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Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.
Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

f_n = nf_1

where

n is the number of the harmonic

f_1 is the fundamental frequency

Here we have:

f_3=512 Hz is the frequency of the 3rd harmonic

So the fundamental frequency is

f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz

And so, the frequency of the 2nd harmonic is:

f_2=2f_1=2(170.7)=341.3 Hz

3 0
4 years ago
Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se
Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}

Camioneta lenta

x_{C} = x_{C,o} +v_{C}\cdot t

Donde:

x_{S,o}, x_{C,o} - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

v_{S,o} - Velocidad inicial de Speedy Sue, medida en metros por segundo.

v_{S}, v_{C} - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

a_{S} - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

t - Tiempo, medido en segundos.

Si conocemos que x_{S} = x_{C}, x_{S,o} = 0\,m, x_{C,o} = 155\,m, v_{S,o} = 30\,\frac{m}{s}, v_{C} =-5\,\frac{m}{s} y a_{S} = -2\,\frac{m}{s^{2}}, encontramos la siguiente función cuadrática:

155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}

-t^{2}+35\cdot t-155 = 0 (Ec. 1)

Las raíces de esta función son:

t_{1}\approx 29.798\,s, t_{2} \approx 5.201\,s

La colisión ocurriría en la raíz positiva más pequeña, es decir:

t \approx 5.201\,s

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: (v_{C,o} = -5\,\frac{m}{s},  x_{C,o} = 155\,m, t \approx 5.201\,s)

x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)

x_{C} = 128.995\,m

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

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Answer:

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