All categories

4 years ago

How much mass does the sun lose through nuclear fusion per second?

Physics

1 answer:

Mekhanik [1.2K]4 years ago

3 0

Answer:

1.10^6 kg of mass per second

Explanation:

All energy lost by the sun comes from nuclear fusion.

Sun loses energy at 2.5*10^{19}J per hour, that is 9*10^{22}J/s

To find the mass lost by the sun in liberation of energy you use the famous Einstein's equation:

$E=mc^2\\\\m=\frac{E}{c^2}=\frac{9*10^{22}}{(3*10^{8}m/s)^2}=1*10^6kg$

hence, the sun liberates 1.10^6 kg of mass per second

You might be interested in

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

A 15-uF capacitor is connected to a 50-V battery and becomes fullycharged. The battery is

Semenov [28]

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

5 0

3 years ago

The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 58-m-diameter loop

dedylja [7]

Answer:

16.87 m/s

Explanation:

To find the speed of the car at the top, when the normal force is equal the gravitational force, we just need to equate both forces:

$N = P$

$m*a_c = mg$

$a_c$ is the centripetal acceleration in the loop:

$a_c = v^2/r$

So we have that:

$mv^2/r = mg$

$v^2/r = g$

$v^2 = gr$

$v = \sqrt{gr}$

So, using the gravity = 9.81 m/s^2 and the radius = 29 meters, we have:

$v = \sqrt{9.81 * 29}$

$v = \sqrt{284.49} = 16.87\ m/s$

The speed of the car is 16.87 m/s at the top.

5 0

3 years ago

What is a scientific model?

avanturin [10]

Answer:

A <u>scientific model</u> is a physical and/or mathematical and/or conceptual representation of a system of ideas, events or processes. Scientists seek to identify and understand patterns in our world by drawing on their scientific knowledge to offer explanations that enable the patterns to be predicted.

3 0

3 years ago

Read 2 more answers

PLZ HELP DUE IN 10MIN!!!!!!!!!The temperature of the water vapor (H2O) inside a pressure cooker increased from 295 K to 395 K. A

madreJ [45]

Did you ever figure it out, bc now I need it lol.

3 0

3 years ago