Why don't asteroids have atmosphere

Physics

2 answers:

EleoNora [17]3 years ago

4 0

Asteroids are too small to have atmospheres

Sladkaya [172]3 years ago

4 0

They are too small an object has to be big enough to have gravity

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1. A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ba

MariettaO [177]

Momentum = (mass) x (velocity)

Original momentum before the hit =

   (0.16 kg) x (38 m/s) this way <==

   =    6.08 kg-m/s this way <==

Momentum after the hit =

   (0.16) x (44 m/s) that way ==>

   = 7.04 kg-m/s that way ==>

Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .
-----------------------------------------------

Change in momentum = impulse.

   Impulse = (force) x (time the force lasted)

13.12 kg-m/s = (force) x (0.002 sec)

(13.12 kg-m/s) / (0.002 sec) = Force

   6,560 kg-m/s² = 6,560 Newtons = Force

    ( about 1,475 pounds ! ! ! )

8 0

2 years ago

A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?

Studentka2010 [4]

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

$W=40 kJ = 40,000 J$ is the work done

$\theta=0^{\circ}$ if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

$d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m$

Learn more about work:

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brainly.com/question/6443626

#LearnwithBrainly

7 0

2 years ago

E fundamental frequency of an open organ pipe corresponds to the middle c (261.6 hz on the chromatic musical scale). the third r

luda_lava [24]

The wavelength of the third resonance of the closed organ pipe is equal to the ratio between the speed of sound and the frequency of the 3rd harmonic:

$\lambda_3 = \frac{c}{f_3}=\frac{343 m/s}{261.6 Hz} =1.31 m$