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Crank
3 years ago
5

Angular diameter of the Sun from the Earth orbit is approximately 32′ (arc minutes), and the solar constant is 1.36kWt/m2 . (a)

What is the Sun’s radius? (b) How much total power does the Sun emit? (c) What is the power flux on the surface of the Sun?
Physics
1 answer:
Burka [1]3 years ago
4 0

Answer:

a) 1392.54 *10^6 m

b) 3.8248284*10^26 W

c) 6.2783674*10^7 W/m^2

Explanation:

from the exercise we have the following values:

angular diameter of sun from earth, phi = 32'

solar constant, S = 1.36 kW/m^2

a. distance of sun from earth, d = 149.6*10^9 m

therefore

phi = D/d

clear D

D = d*phi

so we have

a) diameter of sun = D

D = 1392.54 *10^6 m

b. How much total power does the Sun emit? =P

P = S*4*pi*d^2

P= = 3.8248284*10^26 W

c. power flux on surface of sun = P/pi*D^2

c. power flux on surface of sun  = 6.2783674*10^7 W/m^2

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Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

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Then for this case the relationship becomes: x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m.

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Explanation:

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An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec
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Answer:

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Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

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=>       T_m  =  376.8 ^o C

         

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Answer:

a) F=2.048\times 10^{-7}\ N

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Given:

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  • charge on the raindrops, q=+21\times 10^{-12}\ C
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A)

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F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

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<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

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a=\frac{F}{m}

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