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2 years ago

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The Mass of the Sun is 1.989 × 10^30 kg. The speed of light is 3 x 10^8 m/s. Assuming the Sun is 75% hydrogen, how much Energy c

ould be produced if all of the hydrogen is converted into energy? Use Einstein's famous equation E = mc^2 where E will be in Joules.

1 answer:

bogdanovich [222]2 years ago

8 0

Answer:

$E = 13.428 \times 10^{46} J$

Explanation:

Mass of the Sun = $1.989 \times 10^{30} Kg$

Amount of Hydrogen = 75% of Mass of Sun,

thus mass of Hydrogen (m) in the Sun is,

$m = 1.989 \times 10^{30} \times \frac{75}{100}$

m = $1.492 \times 10^{30}$

Speed of Light (c) = $3 \times 10^{8} m/s$

Thus, energy(E) produced if whole of hydrogen is converted into energy,

E = mc²

$E = 1.492 \times 10^{30} \times (3 \times 10^{8} )^{2}$

$E = 1.492 \times 9 \times 10^{46}$

$E = 13.428 \times 10^{46} J$

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The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 5.4 earth years.

Ad libitum [116K]

Answer:

(a) Radius = 4.6 x 10^11 m

(b) speed = 16.96 km/s

Explanation:

Time period, T = 5.4 earth years

mass of sun, M = 1.989 x 10^30 kg

(a) Let the orbital radius is R.

use the formula of period

$T^2 = \frac{4 \pi^2 R^3}{G M}\\\\\left ( 5.4\times 365\times 24\times 3600 \right )^2 = \frac{4\times3.14\times 3.14\times R^3}{6.67\times10^{-11}\times 1.989\times 10^{30}}\\\\R = 4.6\times 10^{11} m$

(b) Let the speed is v.

$v=\frac{2 \pi\times R}{T}\\\\v=\frac{2\times 3.14\times 4.6\times 10^{11}}{5.4\times 365\times 24\times 3600}\\\\v = 16963.6 m/s =16.96 km/s$

3 0

2 years ago

A 1.5-mole sample of an ideal gas is gently cooled at constant temperature 320 K. It contracts

Alona [7]

Answer:

V₂ = 14.07 L

Explanation:

As this gas is cooled at constant temperature of 320 K, this means that we are on an isothermal process, and according to the 1st law of thermodynamics:

Q = W (1)

And as the temperature is constant, we can use the following expression to calculate the Work done:

W = nRT ln(V₁/V₂) (2)

However, as Q = W, we can replace heat into the above expression and then solve for V₂:

Q = nRT ln(V₁/V₂)

Replacing we have:

1200 = (1.5 * 8.314 * 320) ln(19/V₂)

1200 = 39907.2 ln(19/V₂)

ln(19/V₂) = 1200/3990.72

ln(19/V₂) = 0.3007

19/V₂ = e^(0.3007)

V₂ = 19 / e^(0.3007)

<h2>V₂ = 14.07 L</h2>

Hope this helps

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2 years ago

Question 3 of 10

Fynjy0 [20]

Answer: D

Explanation: a p e x :)

5 0

3 years ago

Read 2 more answers

A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr

Zinaida [17]

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

$m'=9.00 \ g$

or,

$=0.009 \ kg$

mass of wooden block,

$m=1.20 \ kg$

speed,

$s=0.390 \ m$

Coefficient of kinetic friction,

$\mu=0.20$

As we know,

The Kinematic equation is:

⇒ $v^2=u^2+2as$

then,

Initial velocity will be:

⇒ $u=v^2-2as$

$=v^2-2 \mu gs$

On substituting the given values, we get

⇒ $u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}$

$=\sqrt{-1.5288}$

$=1.23 \ m/s$

As we know,

The conservation of momentum is:

⇒ $mu=m'u'$

or,

⇒ Initial speed, $u'=\frac{mu}{m'}$

On substituting the values, we get

⇒ $=\frac{1.20\times 1.23}{0.009}$

⇒ $=\frac{1.476}{0.009}$

⇒ $=164 \ m/s$

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2 years ago

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

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3 years ago