All categories

2 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

Physics

1 answer:

arsen [322]2 years ago

3 0

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

You might be interested in

calculate the energy spent on spraying a drop of mercury of 1 cm radius into 10^6 droplets all of same radius. surface tension o

WARRIOR [948]

Answer:

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

Explanation:

<em>Given:</em>

<em>Radius of mercury = 1cm initially ;</em>
<em>split into $10^{6}$ drops ;</em>

Thus, volume is conserved.

i.e ,

$\frac{4}{3} \pi R_{o}^{3} = 10^{6}*\frac{4}{3} \pi R_{n}^{3}\\R_{n}=\frac{R_{o}}{10^{2}} = \frac{1cm}{100} = 0.01 cm$

Energy of a droplet = $T$ Δ $A$

Where ,

<em>T is the surface tension </em>
<em>ΔA is the change in area</em>

Initial energy $E_{i} = T*A_{i}\\= 0.0035 * 4 *\pi *0.01^{2}\\=4.398*10^{-6}J$

Final energy $E_{f}=10^{6}*T*A_{f}\\=10^{6}*0.0035*4*\pi *(0.0001)^2\\=4.39823*10^{-4}$

∴ .Energy spent on spraying = $=E_{f}-E{i}\\=(439.823-4.39823)*10^{-6}\\=4.3542*10^{-4}J$

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

6 0

2 years ago

A baseball with a mass of 151 g is thrown horizontally with a speed of 40.3 m/s (90 mi/h) at a bat. The ball is in contact with

Ghella [55]

Answer:

A baseball (m= 149g) approaches a bat horizontally at a speed of 40.2 m/s (90 mi/h) and is hit straight back at a speed of 45.6m/s (102mi/h). If the ball is in contact with the bat for a time of 1.10ms, what is the average force exerted on the ball by the bat ? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

Explanation:

Use the impulse equation (a form of Newton's 2nd Law): FΔt = Δ(mv) where Δ means "change in"

The change in momentum is mBB(vf - vi) = (.150 kg)(-46.9 m/s - 40.5 m/s)

Divide this by the time interval and you get F exerted by the bat in Newtons.

Take care.

6 0

3 years ago

Find the weight of an object of mass 5 kg

Elis [28]

Answer:

weight on earth is mg

which is 5*9.8