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Soloha48 [4]
3 years ago
11

An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad

ius a = 2.6 cm and outer radius b = 4.2 cm. The insulating shell is uniformly charged with a volume density of rho = -634 μC/m3. What is λ2, the linear charge density of the insulating shell?
Physics
1 answer:
Anna11 [10]3 years ago
5 0

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

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Answer:

doubled

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2 years ago
A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t
Ganezh [65]

Answer:

The value is \eta  =  54.4 \%

Explanation:

From the question we are told that

    The work input is  W = ( 245 + A ) \  J

     The heat delivered is Q =  (142 + B) \  J

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          \eta  =  \frac{W}{Q_t}

Here  Q_t is the total out energy produce by the heat engine and this is mathematically represented as

           Q_t= Q + W

=>         Q_t=  245 + A + 142 + B

=>         Q_t=  390 + A+B

So

               \eta  =  \frac{245 + A }{390 + A+ B}

=>          \eta  =  0.544

=>          \eta  =  0.544 *100

=>          \eta  =  54.4 \%

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2 years ago
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Answer:

Make sure everything is organized have a planner it can help

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Have your notes

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3 0
3 years ago
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Answer:

W =50 J

Explanation:

given data:

T_h=600 k

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Q_h=100 J

required:

<em>W=??</em>

<u>solution:</u>

║W║=║Q_h║(1-T_L/T_h)

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2 years ago
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