Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.

the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

=  (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π   0_∫^11  rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ   0_∫^11  r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

The answer to the question is;

The plate be left in the furnace for 905.69 seconds.

Explanation:

To solve the question, we have to check the Bi number as follows

Bi =

As the Bi number is > 0.1 we have to account for the variation of temperature with location in the mass.

We perform nonlumped analysis

The relation for heat transfer given by

Y =  

= = 0.3968 = C₁ exp (ζ₁² F₀)  

where

C₁ and ζ₁ are coefficients of a series solution

We therefore look for the values of C₁ and ζ₁ from Bi tables to be

ζ₁ = 0.4801 +(0.26-0.25) (0.5218-0.4801)/(0.3-0.25) ≈ 0.4884 and

C₁ = 0.4801 +(0.26-0.25) (1.0450 - 1.0382)/(0.3-0.25) ≈ 1.03956 and  

This gives the relation

0.3968 = 1.03956 exp (ζ₁² F₀)  

or ζ₁²

where

α = Thermal diffusivity of solid = k/(ρ·c) = = 1.1146×10⁻⁵

c = Specific heat capacity of solid at constant pressure = 550 J/kg·K

ρ = Density of the solid = 7830 kg/m³

=㏑ = -0.9631 from where we have

t =  = 905 seconds.