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Sav [38]
4 years ago
15

The thermal conductivity of a cement is 0.168 British thermal units per foot hour degree Fahrenheit ​[BTU divided by (ft h Super

script o Baseline Upper F )​]. Convert this value in units of watts per meter kelvin ​[Upper W divided by (m Upper K )​].
Physics
1 answer:
Sliva [168]4 years ago
5 0

Answer:

thermal conductivity =  0.2906 W/m.K

Explanation:

thermal conductivity = 0.168 BTU/h.ft⁰F

1 BTU/h.ft⁰F = 1.7295772056 W/m.k

0.168 BTU/h.ft⁰F = 0.168 * 1.7295772056 W/m.k

0.168 BTU/h.ft⁰F = 0.2906 W/m.k

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Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
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A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe
tekilochka [14]

Answer:

Angular momentum, L=6.47\times 10^{-3}\ m

Explanation:

It is given that,

Radius of the axle, r=3.21\ mm=3.21\times 10^{-3}\ m

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

\tau=\dfrac{dL}{dt}

Torque acting on the top is given by :

\tau=F\times r

Here, F is the tension acting on it. Torque acting on the top is given by :

\tau=2F\times r

2T\times r=\dfrac{L}{t}

L=2T\times r \times t

L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32

L=6.47\times 10^{-3}\ m

So, the angular momentum acquired by the top is 6.47\times 10^{-3}\ m. Hence, this is the required solution.

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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