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umka21 [38]
3 years ago
15

En un recipiente de cobre de forma rectangular de aproximadamente 3.5 m largo por 4.5m ancho tiene una temperatura en temporada

de invierno 4 ° C . Calcular su superficie cuando este en verano a una temperatura 40 ° C . ( Cobre 16.6 x 10 ( -6 )
Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

A_f= 15,769 m²

Explanation:

Este es un ejercicio de dilatación térmica,  

          ΔA = (2α) A₀ ΔT

el arrea de recipiente

          A₀ = L A

          A₀ = 3,5 4,5

          A₀=  15,75 m²

el coeficiente de dilatación térmica es alfa = 16,6 10⁻⁶ C⁻¹

calculemos

            ΔA = 2 16,6 10⁻⁶  15,75 ( 40 -4)

            ΔA = 522,9 (36)  10⁻⁶  

            A= 1,88 10⁻²  m2

el cambio de volumen es

          ΔA = A_f – A₀

          A_f = A₀ +  ΔA

          A_f=  15,75 +1,88 10⁻²  

          A_f= 15,769 m²

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An object is placed near a concave mirror having a radius of curvature of magnitude 60 cm. How far should you place the object f
Viefleur [7K]

Answer:

u = 18 cm

Explanation:

given,

radius of curvature = 60 cm

magnification of mirror = 2.5

distance of object  = ?

R = 2 f

f = R/2

f = 60/2 = 30 cm

m = -\dfrac{v}{u}

2.5 = -\dfrac{v}{u}

v = -2.5 u

now,

Using mirror formula

\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}

\dfrac{1}{30} = \dfrac{1}{-2.5u} + \dfrac{1}{u}

\dfrac{1}{30} = \dfrac{0.6}{u}

u = 0.6 x 30

u = 18 cm

distance of object be equal to u = 18 cm

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3 years ago
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Answer:     D

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KATRIN_1 [288]

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3 years ago
Which type of mine involves digging tunnels and shafts deep underground?
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<h3>What is mining?</h3>

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7 0
2 years ago
If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli
Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

There are 2*pi radians in 1 revolution so:

RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

So you need about 30 RPM to keep people from falling out the bottom

7 0
3 years ago
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