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Lilit [14]
4 years ago
11

A silver bar of length 30 cm and cross-sectional area 1.0 cm2 is used to transfer heat from a 100°C reservoir to a 0°C block of

ice.
How much ice is melted per second? (For silver, k = 427 J/s⋅m⋅°C. For ice, Lf = 334 000 J/kg.)
a. 4.2 g/s
b. 2.1 g/s
c. 0.80 g/s
d. 0.043 g/s
Physics
1 answer:
koban [17]4 years ago
4 0

Answer:

d. 0.043 g/s

Explanation:

Formula for rate of conduction of heat through a bar per unit time is as

follows

Q = k A ( t₁ - t₂ ) / L

A is cross sectional area and L is length of rod ,( t₁ - t₂ )  is temperature

difference . Q is heat conducted per unit time

Putting the values in the equation

Q = (427 x 1 x 10⁻⁴ x 100 )/ 30 x 10⁻²

= 14.23 J/s

mass of ice melted per second

= Q / Latent heat of ice

= 14.23 / 334000

= 0.043 g/s

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Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s

In free fall

v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s

The time the rocket is accelerating is 6.75 seconds

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4 years ago
How big is the universe?<br> sorry this is so random
prisoha [69]

Answer:

Explanation: 46 billion light years I think

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6 0
3 years ago
How do professional bicycle riders reduce friction. Name two ways.<br> Thx
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3 years ago
What is made of one type of atom​
umka21 [38]

Any sample of an <em>ELEMENT</em> is made of only one type of atom.

Here are some elements:

-- Hydrogen, Helium, Neon

-- Carbon (lead in a pencil, also diamonds)

-- Oxygen, Nitrogen, Argon (All mixed together in air, but not hooked up with other atoms)  

Other elements you may have heard of:  

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8 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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