I need help please with this question

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

3 years ago

What is the definition of physics

Stella [2.4K]

Answer:

the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.

3 0

2 years ago

How does the orbital speed of an asteroid in a circular solar orbit with a radius of 4.0 AU compare to a circular solar orbit wi

Murrr4er [49]

Answer:

The circular solar orbital speed at 4.0AU is 1/4( one fourth) that at 1.0AU

Explanation:

am = mvr= angular momentum

am4= 4mvt

am1= mvp1

Vt=1/4vp

Vp=4vt

am1= 4mvt

am1=am4

The circular solar orbital speed at 4.0AU is 1/4 (one fourth) that at 1.0AU

6 0

3 years ago

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by $\dfrac{v_i+v_f}2$ , and we know that $v_f=-v_i$ .

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago

Echolocation is based on the use of sound

Anni [7]

Yes echolocation is based on the use of sound and knowing where the sound comes from without having to look for it a lot of soldiers on the battle field need to know how to use echolocation so they can basically not die

5 0

3 years ago

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