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3 years ago

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

Physics

1 answer:

arsen [322]3 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

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

3 years ago

Air expands adiabatically in a piston–cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1

lutik1710 [3]

Answer:

Final temperature is equal to 1291.63°R

Explanation:

given,

p₁ = 100 lb f/in², v₁ = 3.704 ft³/lb, and T₁ = 1000 °R

p₂ = 30 lb f/in² n = 1.4

Δ u = 0.171(T₂ - T₁)

we know for poly tropic process

p vⁿ = constant

p₁ v₁ⁿ = p₂ v₂ⁿ

100 × 3.704¹°⁴ = 30 × v₂¹°⁴

v₂ = 8.753 ft³/lb

work done for poly tropic process

W = $\dfrac{p_1v_1-p_2v_2}{n-1}$

= $\dfrac{100\times 3.704-30\times 8.753}{1.4-1}$

= 269.525 lbf/in².ft³

W = $\dfrac{269.525}{5.40395}$ Btu/lb

= 49.87 Btu/lb

in the piston cylinder arrangement air is expanding acrobatically

Δ q = Δu + w

Δ u = - w

0.171(T₂ - T₁) = -49.87

0.171(T₁ - T₂) = -49.87

0.171 T₂ = 0.171 × 1000 + 49.87

T₂ = 1291.63 °R

Final temperature is equal to 1291.63°R

5 0

3 years ago

The earth is composed of four layers three of these layers are solid and only one is liquid which layer exits in the liquid stat

Vladimir [108]

Answer:

The outer core is a liquid

Explanation:

The outer core is a fluid composed of iron and nickel

7 0

4 years ago

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Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and

Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

$v_f^2=v_i^2+2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

$F_f = \mu_k (mg) \rightarrow$ Frictional Force

$F = ma \rightarrow$ Force by Newton's second Law

Where,

m = mass

a= acceleration

$\mu_k =$ Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

$F_f = F$

$\mu_k mg=ma$

$a = \mu_k g$

Therefore,

$v_f^2=v_i^2+2ax$

$0=v_i^2+2(\mu_k g)x$

Re-arrange to find x,

$x = \frac{v_i^2}{2(-\mu_k g)}$

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0

3 years ago

The acceleration vector of a particle in projectile motion ________.

Alex73 [517]

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

4 0

3 years ago