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3 years ago

E=mc2 sun is losing 4.3 billion kg mass each second ? can you let me know why ?

1 answer:

7 0

Yes I can. You've come to the right place. The reason is because that's where the sun's energy comes from. The sun converts that much mass into energy every second. Then the energy leaves the Sun and shines out into space. A tiny tiny tiny tiny tiny bit of it hits the Earth and it sustains life on Earth.

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Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements

masya89 [10]

Answer:

When the starting and ending points are the same, the total work is zero.

Explanation:

option ( D )correct

A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.

8 0

2 years ago

Find the speed vfinal of the joined cars after the collision. mastering physics

Tanya [424]

<span>Px = 0 Py = 2mV second, Px = mVcosĎ† Py = â€“mVsinĎ† add the components Rx = mVcosĎ† Ry = 2mV â€“ mVsinĎ† Magnitude of R = âš(RxÂ˛ + RyÂ˛) = âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) and speed is R/3m = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) simplifying Vf = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) Vf = (1/3)âš((VcosĎ†)Â˛ + (2V â€“ VsinĎ†)Â˛) Vf = (V/3)âš((cosĎ†)Â˛ + (2 â€“ sinĎ†)Â˛) Vf = (V/3)âš((cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) Vf = (V/3)âš(cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) using the identity sinÂ˛(Ď)+cosÂ˛(Ď) = 1 Vf = (V/3)âš1 + 4 â€“ 2sinĎ†) Vf = (V/3)âš(5 â€“ 2sinĎ†)</span>

6 0

3 years ago

How do you solve <img src="https://tex.z-dn.net/?f=4x%5E%7B3%7D" id="TexFormula1" title="4x^{3}" alt="4x^{3}" align="absmiddle"

Vlad1618 [11]

$Hello$ $There!$

Here's a explanation!

Let's solve your equation step-by-step.

$4x^3=2x^-^1$

$4x^3=\frac{2}{x}$

Step 1: Multiply both sides by x.

$4x^4=2$

$\frac{4x^4}{4} =\frac{2}{4}$

(Divide both sides by 4).

$x^4=\frac{1}{2}$

$x=+(\frac{1}{2} )^(^\frac{1}{4} ^)$

Take the root.

ANSWER!

$x=0.840896$ $Or$ $x=-0.840896$

Hopefully, this helps you!!

$AnimeVines$

8 0

3 years ago

To Calculate Velocity and Average Velocity you can use the same formula as both.

Elanso [62]

Answer:

I believe it's True! Brainliest??

Explanation: Hope you have a great day :)

4 0

2 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

2 years ago