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Yuliya22 [10]
3 years ago
10

6. The two ends of an iron rod are maintained at different temperatures. The amount of heat thatflows through the rod by conduct

ion during a given time interval does notdepend uponA) the length of the iron rod.B) the thermal conductivity of iron.C) the temperature difference between the ends of the rod.D) the mass of the iron rod.E) the duration of the time interval.Ans: DDifficulty: MediumSectionDef: Section 13-27. The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twiceas long and twice the diameter conduct heat between the same two temperatures
Physics
1 answer:
motikmotik3 years ago
4 0

Answer:

20cal/s

Explanation:

Question:

There are two questions. The first one has been answered:

From the formular, Power = Q/t = (kA∆T)/l

the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.

The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).

Second question:

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Solution:

Power = 10cal/s

Power = energy per unit time = Q/t

Where Q = energy

Power = (kA∆T)/l

k = thermal conductivity of iron

A = area

Area = πr^2

r = radius

Diameter = d = 2r

r = d/2

Area = (πd^2)/4

Length = l

∆T = change in temperature

10 = (kA∆T)/l

For a steel rod with length doubled and diameter doubled:

Let Length (L) = 2l

Diameter (D)= 2d

Area = π [(2d)^2]/4 = (π4d^2)/4

Area = 4(πd^2)/4

Using the formula Power = (kA∆T)/l, insert the new values for A and l

Power = [k × 4(πd^2)/4 × ∆T]/2l

Power = [4k((πd^2)/4) ∆T]/2l

Power = [(4/2)×k((πd^2)/4) ∆T]/l

Power = [2k(A) ×∆T]/l = 2(kA∆T)/l

Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l

Recall initial Power = (kA∆T)/l = 10cal/s

And ∆T is the same

2[(kA∆T)/l] = 2 × 10

Power of a steel that has its length doubled and diameter doubled = 20cal/s

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A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
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Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

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The net force can be written as

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A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien
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Explanation:

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   The  mass of the metal rod is  m  = 0.12 \ kg

    The current on the rod is  I  = 4.1 \ A

    The distance of separation(equivalent to length of the rod ) is L   = 6.3 \ m

     The coefficient of kinetic friction is \mu_k  =  0.18

      The kinetic frictional force is  F_k  = 0.212 \ N

     The constant speed is v  = 5.1 \ m/s

Generally the magnetic force on the rod is mathematically represented as  

      F  =  B * I  *   L

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        F_ k  =  B*  I  *  L

=>      B  =  \frac{F_k}{L  *  I  }

=>       B  =  \frac{0.212}{ 6.3   *  4.1   }

=>       B  =  8.20 *10^{-3} \  T

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Calculate the kinetic energy of a car of mass 1.0*10^3kg moving with a velocity of 108km/h<br>​
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Answer:

450000 kg m/s

Explanation:

convert 108km/h to m/s

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108km/h= (108×1000)÷3600

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this should be your answer, hope I was able to help.

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