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Vikentia [17]
3 years ago
7

What type of planets does the asteroid belt orbit?​

Physics
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

Main Asteroid Belt: The majority of known asteroids orbit within the asteroid belt between Mars and Jupiter, generally with not very elongated orbits. The belt is estimated to contain between 1.1 and 1.9 million asteroids larger than 1 kilometer (0.6 mile) in diameter, and millions of smaller ones.

Explanation:

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A _____________ is defined as a push or pull that acts on objects and, in some instances, changes their position.
Shkiper50 [21]
Force !!!!!!!!!!!!!!!!!!!!!!!!!!!
6 0
2 years ago
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A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit
barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

4 0
3 years ago
Plants make food through photosynthesis, a chemical reaction. What are the starting substances of the reaction?
vekshin1
The answer is the 4th option because they don't use oxygen, that's what they produce
3 0
3 years ago
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a student measure the mass of an 8 cm^3 block of brown sugar to be 12.9 g. what is the density of the brown sugar?
schepotkina [342]
Density = mass / volume ;
1 Cubic Centimeter = 0,000001 Cubic Meter
8 cm^3 = 0.000008 m^3
12,9 g = 0,0129 kg


The density is 0,0129 kg/ 0,000008 m^3 = <span><u>1612,5 kg/m^3</u> </span>
4 0
3 years ago
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man
Ne4ueva [31]

Answer:  

A. α = - 1.047 rad/s²  

B. θ = 14.1 rad  

C. θ = 2.24 rev  

Explanation:  

A.  

We can use the first equation of motion to find the acceleration:

\omega_f = \omega_i + \alpha t  

where,  

ωf = final angular speed = 0 rad/s  

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s  

t = time = 3 s  

α = angular acceleration = ?  

Therefore,

0\ rad/s = 3.14\ rad/s + \alpha(3\ s)  

<u>α = - 1.047 rad/s²</u>

B.  

We can use the second equation of motion to find the angular distance:

\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2  

<u>θ = 14.1 rad</u>

C.  

θ = (14.1 rad)(1 rev/2π rad)  

<u>θ = 2.24 rev</u>

6 0
3 years ago
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