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Natasha2012 [34]
3 years ago
14

A carpenter builds an exterior house wall with a layer of wood 3.1 cm thick on the outside and a layer of Styrofoam insulation 2

.3 cm thick on the inside wall surface. The wood has k = 0.080W/(m⋅K), and the Styrofoam has k = 0.010 W/(m⋅K). The interior surface temperature is 19.0 ∘C, and the exterior surface temperature is -15.0 ∘C.A.)What is the temperature at the plane where the wood meets the Styrofoam?
B.)What is the rate of heat flow per square meter through this wall?
Physics
1 answer:
lapo4ka [179]3 years ago
6 0

Answer: T = -0.213°c

Heat flow = 38.2 w/m²

Explanation:

given data:

Wood = 0.031m thick

Styrofoam = 0.023m thick

Wood = k ( 0.080W/m.k)

Styrofoam = k (0.010W/m.k)

Temperature of wood = -15°c

Temperature of styrofoam = 19°c

(a) heat flow in wood and styrofoam must be equal

KAdT/ L = KAdT/ L

0.080 * A * ( T - ( -15) ) / 0.031 = 0.010 * A * ( T - ( 19°c) ) / 0.023

T = -0.213°c

(b) heat flow

H = KAdT / L

H = 0.080 * A * ( -0.213°c - ( -15°c) ) / 0.031

= 38.2 w/m²

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8_murik_8 [283]

Answer:

Ve(m) = sqrt (19.2/m)

Ve(0.35) = 7.407 m/s

Explanation:

Given:

- The ball has a mass = m

- The entry speed of the ball is Vi = Ve

- The final speed of the ball Vf = 0.5*Ve

- The constant frictional force on ball due to hay is F = 6 N

- The thickness of hay-stack is s = 1.2 m

- Assume the throw is in horizontal direction and neglect gravity forces

Find:

Derive an expression for the typical entry speed as a function of the inertia of the ball

What is the typical entry speed if the ball has an inertia of a 0.35 kg?

Solution:

- To determine the entry speed as a function of inertia we will use third equation of motion as follows:

                               Vf^2 = Vi^2 + 2*a*s

Where, a is acceleration of the ball through hay stack. We will use Newton's Law of motion to determine this:

                               F_net = m*a

The only force acting on the ball in its journey through hay-stack is the frictional force F:

                               - F = m*a

                                a = -F/m

- Input all the quantities in the third equation of motion:

                                (0.5Ve)^2 = Ve^2 - 2*F*s / m

                                0.75Ve^2 = 2*F*s / m

                                Ve = sqrt (8*F*s/3*m)

Plug in values:

                                Ve(m) = sqrt (8*6*1.2/3*m)

                                Ve(m) = sqrt (19.2/m)

- The entry speed for the inertia of the ball m = 0.35 kg is:

                                Ve(0.35) = sqrt(19.2/0.35)

                                Ve(0.35) = 7.407 m/s

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