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klemol [59]
3 years ago
14

If you weigh 690n on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and

a diameter of 19.0 km ?]
Physics
1 answer:
NeTakaya3 years ago
5 0
-- If you weigh 690N on the Earth, then your mass is about

           (690N) / (9.8 m/s²) = 70 kg .

-- If your mass is 70 kg then your weight on that neutron star would be

              G  M₁  M₂  /  R²

         =  (6.67 x 10⁻¹¹ N m²/kg²) (70 kg) (1.989 x 10³⁰ kg) / (19,000 m)²

         =  (6.67 x 10⁻¹¹ x 70 x 1.989 x 10³⁰) / (3.61 x 10⁸)        N

         =  (6.67 x 70 x 1.989 x 10¹⁹)  /  (3.61 x 10⁸)          N

         =    2.57 x 10¹³    Newtons .

  (about  2,890,000,000 tons !)
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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
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Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

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Setting goals that match your lifestyle helps improve chances that they will be reached.
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Answer:

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How does electroscope detect type of charge ?
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Answer:

An electroscope is an early scientific instrument used to detect the presence of electric charge on a body. It detects charge by the movement of a test object due to the Coulomb electrostatic force on it. The amount of charge on an object is proportional to its voltage.

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Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
kobusy [5.1K]

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

  • V = Linear speed
  • W = Angular speed
  • r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2\pi/T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x 10^{-4} Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x  10^{-4} x 149.6 x 10^{6}

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4\pi ^{2}r^{3}) / GT^{2}

M = (4 x 9.8696 x 3.35 x 10^{24}) / (6.67 x 10^{-11} x 7.68 x 10^{11}<em>)</em>

<em>M = 1.32 x </em>10^{26} / 51.226

M = 2.58 x 10^{24} Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

#SPJ1

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2 years ago
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