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MA_775_DIABLO [31]
3 years ago
7

The specific heat of Aluminum is 0.9 J/g K. The specific heat of Copper is 0.39 J/g K. If samples of equal mass of both Aluminum

and Copper are heated up to 100°C and then dropped in a cold water bath. Compare the heat lost by the two samples.
Physics
2 answers:
AleksandrR [38]3 years ago
6 0
<span> <span><span> <span> Answer:The Aluminum loses a little more than twice the heat of the Copper.Explanation:<span>
Since specific heat is part of the equation. A smaller specific heat will create a smaller heat gain or loss. </span>
<span>Hope this helped!!!!</span></span> </span> </span></span>
viva [34]3 years ago
4 0

Answer:

Heat loss by  aluminium will be more as compare to copper.

Explanation:

We know that heat lost given as follows

Q=mC_p\Delta T

Given that mass of aluminium and copper are same so lets take 1 g.

Lets take the temperature of cold water is 10°C

For aluminium

Q=mC_p\Delta T

Q=1\times 0.9\times 90

Q= 81 J

For copper

Q=mC_p\Delta T

Q=1\times 0.39\times 90

Q= 35.1 J

So from above we can say that heat loss by  aluminium will be more as compare to copper.

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T = 2 \pi \sqrt {l / g}

In this way, by bringing their legs closer they reduce the distance and make the period shorter.

7 0
3 years ago
Una persona cierra una puerta de 1 metro de ancho aplicando una fuerza de 40 [N], perpendicular a ella, a 90 [cm] de su eje de r
Damm [24]

Answer:

El módulo del torque aplicado es 36 Nm

Explanation:

En los movimientos rotatorios, la cantidad de fuerza aplicada no depende de la acción gravitacional sino del momento inercial, que es el equivalente angular de la inercia (masa) y representa la resistencia que un objeto ofrece al rotar alrededor de su eje. Cuando un cuerpo rígido rota alrededor de su eje debe considerarse , además de la masa, el radio de giro ya que estos dos factores determinan la resistencia del cuerpo a los cambios de movimiento rotatorio a través de un eje determinado.

De esta manera, se llama torque o momento de una fuerza a la capacidad de dicha fuerza para producir un giro o rotación alrededor de un punto.

En muchas ocasiones el punto de aplicación de la fuerza no coincide con el punto de aplicación en el cuerpo. En este caso la fuerza actúa sobre el objeto y su estructura a cierta distancia, mediante un  elemento que traslada esa acción de esta fuerza hasta el objeto. Entonces, el momento de una fuerza  es, matemáticamente,  igual al producto de la intensidad de la fuerza (módulo) por la distancia desde el punto de aplicación de la fuerza hasta el eje de giro:

M=F*d*sen θ

donde F es la fuerza en Newton (N), d la distancia en metros (m), θ el ángulo que forma la fuerza con el objeto al cual se le aplica la fuerza y M el momento, que se mide en Newton por metro (Nm).

En este caso:

  • F= 40 N
  • d= 90 cm= 0.9 m (siendo 100 cm= 1 m)
  • θ= 90° ya que la fuerza se aplica de forma perpendicular. Entonces sen θ= sen 90= 1

Reemplazando:

M=40 N*0.9 m* 1

Resolviendo:

M= 36 Nm

<u><em>El módulo del torque aplicado es 36 Nm</em></u>

3 0
3 years ago
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