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1 month ago

The total conversion of 1.00 kilogram of the Sun's mass into energy yields?

Physics

1 answer:

Yuliya22 [3.3K]1 month ago

5 0

The well-known equation... E = m c²... does not address the origin of the mass involved.

Converting 1 kg of any mass entirely into energy generates

(1kg) · (c²) Joules of energy.

E = (1 kg) · (c²) = (1 kg) · (299,792,458 m/s)²

E = 8.9876 x 10¹⁶ Joules

To simplify, this equates to the energy needed to keep a 100-watt light bulb illuminated for about 10,402,259,010 days.

(This is roughly 28.5 million years, based on the current understanding of days and years.)

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An airplane flies with a velocity of 55.0 m/s [35o N of W] with respect to the air (this is known as air speed). If the velocity

ValentinkaMS [3465]

V - wind speed;
53° - 35° = 18°
v² = 55² + 40² - 2 · 55 · 40 · cos 18°
v² = 3025 + 1600 - 2 · 55 · 40 · 0.951
v² = 440.6
v = √440.6
v = 20.99 ≈ 21 m/s
Conclusion: The wind speed calculates to 21 m/s.

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1 month ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

inna [3103]

Answer:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$ ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + $z^{2}$ k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + $v^{2}$ k

The normal vector N is computed as

$N = S_{u}XS_{v}$

Where:

$S_{u} = =$

$S_{v} = =$

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, $v^{2}$ >. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + u $v^{2}$

Thus

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu$ ≈ 3077.34

8 0

1 month ago

A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

Given 52.92 g of metal in the sample, the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0

2 months ago

A plasma is a gas of ionized (charged) particles. When plasma is in motion, magnetic effects "squeeze" its volume, inducing inwa

Softa [3030]

Answer:

The density of volume charge is $\rho = nq$

The density of surface charge is $\sigma = n^{\frac{2}{3} } q$

Explanation:

The question states that

The radius measures R

The length is L

The speed is v

The ion count per unit volume is n

The charge per ion is q

The surface thickness of the cylinder is $n^{\frac{1}{3} }$

The volume charge density is mathematically expressed as

$\rho = nq$

The surface charge density is mathematically expressed as

$\sigma = \rho n^{\frac{1}{3} }$

substituting for $\rho$

$\sigma = n * n^{\frac{1}{3} } q$

$\sigma = n^{\frac{2}{3} } q$

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1 month ago

The total conversion of 1.00 kilogram of the Sun's mass into energy yields​?

The total conversion of 1.00 kilogram of the Sun's mass into energy yields?