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sasho [114]
3 years ago
6

Please help thanks ;)

Physics
2 answers:
Nikitich [7]3 years ago
8 0

Answer:b speed with displacement

Explanation:

notsponge [240]3 years ago
4 0

Answer:

B. Velocity is speed with displacement

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
Based on what you learn about jovian moons by watching the videos or reading your textbook, what types of evidence for recent or
Valentin [98]

Answer:

b. craters, river valleys feeding into surface lakes of very cold liquids

Explanation:

  • Jovian moons are the four largest satellites like the moon of the Jupiter ie the Lo, Europa, Ganymede, Callisto and were first seen by the galileo. They are amiugly the largest moons with radii larger than the dwarf planet.  
  • Lo has more than 400 active volcanoes and dotted more than 100 mountains and has an extremely thin atmosphere made up of sulfur dioxide. The Europa has deep oceans of liquid water, and the layer of ice, and are characteristic of the tidal heating.  
  • <u>While the surface of Callisto is heavily cratered and has salty liquid water.</u>
7 0
3 years ago
Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0
3 years ago
What is the kinetic energy of an automobile with a mass of 1252 kg traveling at a speed of 12 m/s?
torisob [31]
KE = 1/ 2 * 1252 * 144
 as  KE = 1/2 * m * v ^2
 = 90144 J



4 0
3 years ago
A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:
Ilia_Sergeevich [38]

Answer:

  1. 6.198 m/s²
  2. 4.48 s
  3. 453.77 m

Explanation:

5 0
3 years ago
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