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maw [93]
3 years ago
6

As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant

Physics
1 answer:
Alika [10]3 years ago
4 0

Answer:

Isolated or Closed system, both are correct

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Calculate What is the atomic number of<br> a sodium atom that has 11 protons and 12 neutrons
miv72 [106K]

Answer:

11 because the number of protons is the atomic humber

Explanation:

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How can you describe the motion of an object in a race?
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You can describe the motion of an object by its position, speed, direction, and acceleration
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3 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

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3 0
3 years ago
The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the
Margaret [11]

Answer:

v=2019.09\ m.s^{-1}

Explanation:

Given:

  • time taken by the sun to complete one revolution, t=601.2\ hr
  • radial distance of the sunspot, r=6.955\times 10^8\ m.s^{-1}

<u>Therefore, angular speed of rotation of sun:</u>

\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}

<u>Now the tangential velocity of the sunspot can be given by:</u>

v=r.\omega

v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}

v=2019.09\ m.s^{-1}

4 0
3 years ago
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
Fynjy0 [20]

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

\Delta L=\dfrac{PL}{AE}

A=\dfrac{PL}{\Delta LE}

A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

4 0
3 years ago
Read 2 more answers
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