If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
1 answer:
Answer:
Average density of Sun is 1.3927 .
Given:
Radius of Sun = 7.001 × km = 7.001 ×
cm
Mass of Sun = 2 × kg = 2 ×
g
To find:
Average density of Sun = ?
Formula used:
Density of Sun =
Solution:
Density of Sun is given by,
Density of Sun =
Volume of Sun =
Volume of Sun =
Volume of Sun = 1.436 ×
Density of Sun =
Density of Sun = 1.3927
Thus, Average density of Sun is 1.3927 .
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<h2>Answer:</h2>
(e) halved
<h2>Explanation:</h2>The electrical enery (E) stored in a capacitor is related to its capacitance (C) and potential difference (V) as follows;
E = x C x
------------------------(i)
Also, the capacitance (C) of a capacitor consisting of parallel plates is related to the area (A) of the plates and distance (d) between the plates as follows;
C = A x ε₀ / d ------------------------(ii)
Where;
ε₀ is the permittivity of free space.
Substituting equation (ii) into equation (i) gives;
E = x A x ε₀ / d x
--------------------(iii)
From equation(iii)
When the potential difference (V) is constant, then the electrical energy (E) stored is <em>inversely </em>proportional to the distance between the plates. i.e
E = k / d ----------------(iv)
Where;
k = proportionality constant = x A x ε₀ x
(which is the product of all constants)
Therefore from equation (iv);
=> E₁ x d₁ = E₂ x d₂ ---------------------------(v)
Where;
E₁ and E₂ are the initial and final values of the electrical energy stored.
d₁ and d₂ are the initial and final values of the distance between the plates.
<em>So, when the distance is doubled, i.e.</em>
d₂ = 2 x d₁
<em>Substitute the value of d₂ into equation (v) to give;</em>
=> E₁ x d₁ = 2 x d₁ x E₂
<em>Divide through by d₁ to give;</em>
=> E₁ = 2 x E₂
<em>Make E₂ subject of the formula</em>
=> E₂ = x E₁
Therefore, the electrical energy stored in the capacitor will be halved.
Answer:
= - 26.31 kJ
Explanation:
we know that number of moles is calculated as
= 0.00476 mol
Heat absorbed by calorimeter
= 26.87 kJ
Enthalpy of combustion
= - 55290.12 kJ/mol
Negative sign shows that the heat is released
The balanced reaction
ΔHc = ΔU + Δng (RT)
-55290.12 = ΔU + (12 - 12) *(RT)
= - 26.31 kJ