Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 33, Problem 49AP

(a)

To determine

The radius of the hemisphere.

(a)

Expert Solution
Check Mark

Answer to Problem 49AP

The radius of the hemisphere is 0.161m .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the total mass is,

m=mc+4mk

Here,

mc is the mass of cat.

mk is the mass of the kitten.

Substitute 5.50kg for mc and 0.800kg for mk in the above equation to find the value of m .

m=5.50kg+4(0.800kg)=8.7kg

The formula to calculate the mass of the hemisphere is,

m=(23πr3)ρ (1)

Here,

r is the radius of the hemisphere.

ρ is the density of the hemisphere.

Substitute 8.7kg for m and 990kg/m3 for ρ in equation (1) to find the value of r .

8.7kg=(23πr3)(990kg/m3)r=(8.7kg)×3(990kg/m3)2π3=0.161m

Conclusion:

Therefore, the radius of the hemisphere is 0.161m .

(b)

To determine

The area of the curved surface.

(b)

Expert Solution
Check Mark

Answer to Problem 49AP

The area of the curved surface is 0.163m2 .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the area is,

A=2πr2

Substitute 0.161m for r in the above equation to find the value of A .

A=2π(0.161m)2=0.163m2

Conclusion:

Therefore, the area of the curved surface is 0.163m2 .

(c)

To determine

The power emitted by the cats.

(c)

Expert Solution
Check Mark

Answer to Problem 49AP

The power emitted by the cats is 76.8W .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the power emitted is,

P=σAT4

Here,

σ is the Stephan’s Boltzmann constant.

T is the temperature of the surroundings.

Substitute 5.54×10 8W/m2K4  for σ , 31.0°C for T and 0.163m2 for A in the above equation to find the value of P .

P=(5.54×10 8W/m2K4 )(0.163m2)(31.0°+273K)4=76.8W

Conclusion:

Therefore, the power emitted by the cats is 76.8W .

(d)

To determine

The intensity of radiation at the surface.

(d)

Expert Solution
Check Mark

Answer to Problem 49AP

The intensity of radiation at the surface is 470W/m2 .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the intensity of radiation is,

I=PA

Substitute 76.8W for P and 0.1634m2 for A in the above equation to find the value of I .

I=76.8W0.1634m2=470W/m2

Conclusion:

Therefore, the intensity of radiation at the surface is 470W/m2 .

(e)

To determine

The amplitude of electric field in the electromagnetic wave.

(e)

Expert Solution
Check Mark

Answer to Problem 49AP

The amplitude of electric field in the electromagnetic wave is 595V/m .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the amplitude of the electric field is,

E=2μcI

Here,

I is the intensity of wave.

c is the speed of the light.

μ is the permeability of vacuum.

Substitute 470W/m2 for I , 4π×107 for μ and 3×108m/s for c in the above equation to find the value of E .

E=2(4π×107)(3×108m/s)(470W/m2)=595V/m

Thus, the amplitude of electric field in the electromagnetic wave is 595V/m .

Conclusion:

Therefore, the amplitude of electric field in the electromagnetic wave is 595V/m .

(f)

To determine

The amplitude of magnetic field in the electromagnetic wave.

(f)

Expert Solution
Check Mark

Answer to Problem 49AP

The amplitude of magnetic field in the electromagnetic wave is 1.98μT .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the amplitude of the magnetic field is,

B=Ec

Substitute 595V/m for E and 3×108m/s for c in the above equation to find the value of B .

B=595V/m3×108m/s=1.98×106T×106μT1T=1.98μT

Conclusion:

Therefore, the amplitude of magnetic field in the electromagnetic wave is 1.98μT .

(g)

To determine

The total power radiated by the family of cat.

(g)

Expert Solution
Check Mark

Answer to Problem 49AP

The total power radiated by the family of cat is 119W .

Explanation of Solution

Given info: The weight of the black cat is 5.50kg , the weight of the kittens is 0.800kg , the temperature of the hemisphere is 31.0°C , the emissivity of the black cat is 0.970 and the uniform density is 990kg/m3 .

Explanation:

The formula to calculate the mass of the hemisphere is,

mk=(23πrk3)ρ

Here,

rk is the radius of the hemisphere covered by kittens.

ρ is the density of the hemisphere.

Substitute 0.800kg for mk and 990kg/m3 for ρ in equation (1) to find the value of rk .

0.800kg=(23πrk3)(990kg/m3)rk=(0.800kg)3(990kg/m3)2π3=0.156m

The formula to calculate the power radiated by the kittens is,

Pk=4σ(2πrk2)T4

Here,

σ is the Stephan’s Boltzmann constant.

T is the temperature of the surroundings.

Substitute 5.54×10 8W/m2K4  for σ , 31.0°C for T and 0.15m for rk in the above equation to find the value of Pk .

Pk=4×(5.54×10 8W/m2K4 )(2π(0.15m)2)(31.0°+273K)4=29.3W

The formula to calculate the mass of the hemisphere is,

mc=(23πrc3)ρ

Here,

rc is the radius of the hemisphere covered by kittens.

ρ is the density of the hemisphere.

Substitute 5.5kg for mc and 990kg/m3 for ρ in equation (1) to find the value of rc .

5.5kg=(23πrc3)(990kg/m3)rc=(5.5kg)3(990kg/m3)2π3=0.29m

The formula to calculate the power radiated by the cat is,

Pc=σ(2πrc2)T4

Here,

σ is the Stephan’s Boltzmann constant.

T is the temperature of the surroundings.

Substitute 5.54×10 8W/m2K4  for σ , 31.0°C for T and 0.29m for rc in the above equation to find the value of Pc .

Pc=(5.54×10 8W/m2K4 )(2π(0.31m)2)(31.0°+273K)4=26.4W

The formula to calculate the total power radiated by the family of cat is,

PT=e(2(Pc+PK)+(Pc+PK)6)

Substitute 0.98 for e , 28.638W for Pc and 29.3W for Pk in the above equation to find the value of PT .

PT=0.98(2(28.638W+29.3W)+(28.638W+29.3W)6)=119W

Conclusion:

Therefore, the total power radiated by the family of cat is 119W .

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Chapter 33 Solutions

Physics for Scientists and Engineers

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