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Kitty [74]
3 years ago
10

How hot is the sun? First answer gets brainliest. :)

Physics
2 answers:
Andreyy893 years ago
5 0
Inner (Core) Temperature = 1.5 crore degree celsius
Surface temp. = 6 k degree celsius

Hope this helps!
AleksAgata [21]3 years ago
3 0
<span>5,778 K is how hot the sun is. ( ty will be receiving my brainliest now :) </span>
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kinetic energy to gravitational potential energy

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I need help. which interaction does not take place due to field forces
Advocard [28]

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Explanation:

Answer D. Interaction between a pen and paper while you write

7 0
3 years ago
A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string i
Firdavs [7]

Answer:

a  

The tension in the string is  T = 0.85 N

 b

 The new balance reading is  M_b  =  885.86 g

Explanation:

From the question we are told that

    The mass of the beaker of water is  m = 875 .0g

     The diameter of the copper ball is  d = 2.75 cm = \frac{2.75}{100} = 0.0275 m

There are two forces acting on the copper ball

   The first is the Buoyant force of the water pushing it up which is mathematically represented as

                     F = \rho V g

Where \rho is the density of water which has value of  \rho = 1000 kg/m^3

            g is the acceleration due to gravity g= 9.8 \  m/s^2

          V is the volume of water displaced by the copper ball  which is mathematically evaluated as

                             V = \frac{4}{3}  \pi r^3

The radius r is  r = \frac{d}{3}  = \frac{0.0275}{2} = 0.01375 m

Substituting value  

                        V = \frac{4}{3} * 3.142 * (0.01375)^3

                            V = 1.08 9 *10^{-5 } m^3  

   Substituting for  F

              F = 1000 * 1.089 *10^{-5} * 9.8

               F = 0.1067 N      

     The second force is the weight of the copper ball which is mathematically represented as

       W_c = mg

Now m is the mass which can be mathematically evaluated as

        m =  \rho_c * V

Where  is the density of copper with  value of  \rho_c = 8960 kg /m^3

So      m = 8960 * 1.089 *10^{-5}

         m = 0.0976

So the weight of copper is  

             W_c = 0.09756  *  9.8

            W_c = 0.956 N

Now the tension the string would be mathematically evaluated as

            T = W_c - F

So        T = 0.956 - 0.1067

           T = 0.85 N

From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance

Now the mass equivalent of this weight is mathematically evaluated as

             m_z = \frac{1.0645 }{9.8 }

             m_z = 0.01086 kg

Converting to grams  

                     m_z = 10.86 g

So the new balance reading is  

                  M_b  =  875.0 +10.86

                  M_b  =  885.86 g

5 0
3 years ago
Which molecule has a central carbon with four different components around it
sasho [114]
Methane


The many covalent bonds between the atoms in hydrocarbons store a great amount of energy, which releases when these molecules burn
3 0
3 years ago
In an oscillating LC circuit, the maximum charge on the capacitor is qm. Determine the charge on the capacitor, q, and the curre
Katarina [22]

Answer:

a. q = qm/√2 b. qm/√(2LC)

Explanation:

a. Charge on the capacitor

Let U₁ = energy in inductor and U₂ = energy in capacitor and U = total energy in circuit.

So, U₁ + U₂ = U.

Since the energy is shared equally between the capacitor and inductor, U₁= U₂,so

2U₂ = U

and U₂ = U/2

Now U₂ = q²/2C where q is the charge on the capacitor with capacitance C and

U = (qm)²/2C where qm is the maximum charge on the capacitor.

Since U₂ = U/2,

Substituting the values for U₂ and U, we have

q²/2C = [(qm)²/2C]/2

q² = (qm)²/2

taking square-root of both sides, we have

q = qm/√2

b. The current in the inductor

Since the energy in the capacitor equals the energy in the inductor,

1/2LI² = 1/2q²/C where L is the inductance of the inductor and I the current through it.

I² = q²/LC

taking square-root of both sides, we have

I = q/√LC

substituting the value of q from above, we have

I = qm/√2 ÷ √LC

I = qm/√(2LC)

6 0
3 years ago
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