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Otrada [13]
2 years ago
14

Assume that the thickness of the tissue between the interior and the exterior of the body is =3 cm and that the average area thr

ough which conduction can occur is =1.5 m2 . Find heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2 0C. Coefficient of thermal for tissue without blood is = 18 × 10−2 Cal /m-hr- 0C.
Physics
1 answer:
Westkost [7]2 years ago
7 0

The heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.

<h3>What is heat conduction?</h3>

Heat conduction is the method of transfer of heat by the direct contact of the two or more solid bodies.

Given is the thickness of tissue x = 3cm =-0.03m, Area of conduction A = 1.5 m², Change in temperature of skin ΔT =2°C, and coefficient of thermal for tissue without blood is k = 18 × 10⁻² Cal /m-hr- °C

The conduction heat rate in hr is

Q =kAΔT /x

Q =18 × 10⁻² x 1.5 x 2 /0.03

Q =18 Joule / hr

Thus, the heat transferred by conduction per hour, if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.

Learn more about conduction.

brainly.com/question/12136944

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A thin ball shell is shaped like a hollow ball with a radius of 42.8 cm. The material is a conductor with charge 0.8 nC A point
Gala2k [10]

Answer:

The Electric field will be 225.92 N/C

Explanation:

Given :

  • Radius of the hollow sphere R=42.8 cm
  • Initial charge on the conducting sphere is Q_1=0.8\times10^{-9}\ \rm C
  • Magnitude of the point charge q_2=-5.4\times10^{-9}\ \rm C

We know that the flux of the electric field through a close surface cab be calculated by using Gauss Law which is

\int EdA=\dfrac{Q_{in}}{\epsilon_0}\\\\E\times4\pi\times 0.428^2=\dfrac{(0.8-5.4)\times10^{-9}}{\epsilon_0}\\\\E=225.92\ \rm N/C

Hence the Electric Field is calculated.

The answer founded out by you seems to be correct and very close to the exact answer and the concept you have used in your answer is also correct.

5 0
3 years ago
A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the t
aliina [53]

The question seems a bit incomplete. The question should be as follow:

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder. What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g= 9.8 m/s2.)

Answer:

45.08N

Explanation:

The question involves moment topic. Note that the equation of moment, M is

Moment, M = Force, F x Perpendicular distance from the turning point, d

M = F x d

Moment involves turning movement along the pivot. in this case, we have 2 pivots. The first one the attached near the rope, while the other is the crank.

To raise the bucket, minimum force required must be able to provide at least the same moment as the moment due to the bucket of water. i.e.

Moment due to bucket = moment due to force on the crank

First find the moment due to bucket,

Mb = F (weight) x d (radius of cylinder on top of well)

   =  (23 x 9.8) x 0.05

   = 11.27 Nm

Next, Find the moment due to force on the crank. Note that the force is the minimum required force for this equation.

Mc = F (min Force) x d (radius of crank)

     = F x  0.25  = 0.25F

Now we get a simple equation of

11.27 = 0.25F

Using Algebra, we'll get the minimum Force, F:

F = 11.27 / 0.25

  = 45.08 N

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4 years ago
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I got a question is it possible to make sunglasses that let you look at the sun without going blind?
AleksandrR [38]

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4 years ago
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uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work i
Lesechka [4]

Answer:

(a) The work done is 0.05 J

(b) The  force will stretch the spring by 3.8 cm

Explanation:

Given;

work done in stretching the spring from 30 cm to 45 cm, W = 3 J

extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m

The work done is given by;

W = ¹/₂kx²

where;

k is the force constant of the spring

k = 2W / x²

k = (2 x 3) / (0.15)²

k = 266.67 N/m

(a) the extension of the spring, x = 37 cm - 35 cm = 2 cm = 0.02 m

work done is given by;

W = ¹/₂kx²

W = ¹/₂ (266.67)(0.02)²

W = 0.05 J

(b) force = 10 N

natural length L = 30 cm

F = kx

x = F / k

x = 10 / 266.67

x = 0.0375 m

x = 3.75 cm = 3.8 cm

Thus a force of 10 N will stretch the spring by 3.8 cm

7 0
4 years ago
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